Samacheer Kalvi Class 12 Physics Solutions Chapter 3 Magnetism and Magnetic Effects of Electric Current

Properties	Dia	Para	Ferro
Magnetic susceptibility	Negative	Positive and small	Positive and large
Relative permeability	Slightly less than unity	Greater than unity	Large
Nature of magnetic field lines	Repelled or expelled from diamagnetic materials	Attracted into the paramagnetic materials	Strongly attracted to the ferromagnetic materials
Susceptibility	Temperature independent.	Inversely proportional to temperature.	Inversely proportional to temperature.
Examples	Bismuth, copper, and water, etc.	Aluminum, platinum, chromium, etc.	Iron, Nickel, and Cobalt

In simple words: These materials differ in how they react to a magnetic field, showing different levels of attraction or repulsion. This is seen in their magnetic susceptibility, how much magnetic field they let through, and their temperature dependence.

🎯 Exam Tip: For each material type (dia-, para-, ferro-), remember the key characteristics: sign/magnitude of susceptibility, relative permeability value, and examples. Focus on how strongly and in what direction they are magnetized.

 

Question 19. What happens to the domains in a ferromagnetic material in the presence of an external magnetic field?
Answer: When an external magnetic field is applied to a ferromagnetic material, two main things happen to its magnetic domains. First, the domains whose magnetic moments are already aligned parallel to the external field will grow larger in size, taking over space from less favorably oriented domains. Second, the other domains that are not parallel to the field will rotate so that their magnetic moments align themselves with the direction of the applied field. This alignment makes the material strongly magnetized. This process is how a ferromagnetic material becomes a magnet.
In simple words: When you put a ferromagnet in a magnetic field, the tiny magnetic areas (domains) that already point the same way get bigger, and other areas turn to match the field, making the material strongly magnetic.

🎯 Exam Tip: Remember that domain growth and domain rotation are the two primary mechanisms by which ferromagnetic materials magnetize in an external field.

 

Question 20. How is a galvanometer converted into (i) an ammeter and (ii) a voltmeter?
Answer:
1. To convert a galvanometer into an ammeter of a specific range, a suitable low resistance, called a shunt, is connected in parallel with the galvanometer. This shunt allows most of the current to bypass the galvanometer, protecting it from high currents and enabling it to measure a larger total current.
2. To convert a galvanometer into a voltmeter, a suitable high resistance (R_h) is connected in series with the galvanometer. This high resistance limits the current flowing through the galvanometer and allows it to measure a larger potential difference (voltage) across two points in the circuit. This makes it safe for the galvanometer while expanding its measurement range.
In simple words: You turn a galvanometer into an ammeter by adding a small resistor next to it (in parallel). You turn it into a voltmeter by adding a big resistor in line with it (in series).

🎯 Exam Tip: The key difference in conversion is the type of resistance (low/high) and connection (parallel/series) – low parallel resistance for ammeter, high series resistance for voltmeter.

III. Long Answer Questions:

 

Question 1. Discuss Earth’s magnetic field in detail.
Answer:
1. William Gilbert first suggested that Earth acts like a massive bar magnet. However, the Earth's core is too hot for permanent magnetism.
2. Other theories suggest Earth’s magnetic field comes from hot, flowing liquid iron in its outer core, which creates electric currents. This is known as the geodynamo effect.
3. The study of Earth's magnetic field is called geomagnetism or terrestrial magnetism.
4. To fully describe Earth’s magnetic field at any place, we need three main quantities:
* **Magnetic Declination (D):** This is the angle between the magnetic north (where a compass points) and the true geographic north.
* **Magnetic Dip or Inclination (I):** This is the angle that Earth’s total magnetic field lines make with the horizontal surface.
* **Horizontal Component of Earth’s Magnetic Field (\( B_H \)):** This is the horizontal part of the total magnetic field. The total magnetic field is \( B_E \).
5. The vertical component \( B_V = B_E \sin I \) and the horizontal component \( B_H = B_E \cos I \). From this, we can derive \( \tan I = \frac{B_V}{B_H} \).
6. At the magnetic equator, the inclination \( I = 0^\circ \), so \( B_V = 0 \). The field is entirely horizontal. At the magnetic poles, the inclination \( I = 90^\circ \), so \( B_H = 0 \). Here, the field is entirely vertical.
In simple words: Earth has its own magnetic field, like a huge magnet. We describe it using three things: how far off magnetic north is from true north (declination), how much the field dips into the ground (inclination), and how strong the field is sideways.

🎯 Exam Tip: Remember the three key elements that define Earth's magnetic field at any point: magnetic declination, magnetic inclination (dip angle), and the horizontal component of the magnetic field.

 

Question 2. Deduce the relation for the magnetic field at a point due to an infinitely long straight conductor carrying current.
Answer:Let's find the magnetic field at a point P due to a very long, straight wire carrying current I.
1. Imagine a long straight wire, NM, carrying a current I.
2. Pick a small segment of the wire, \( dl \), at a distance \( l \) from a point O on the wire.
3. Consider a point P at a perpendicular distance 'a' from the point O.
4. According to Biot-Savart's law, the magnetic induction \( d\overrightarrow{B} \) at point P due to this small segment \( dl \) is given by: \( \overrightarrow{dB} = \frac{\mu_0}{4\pi} \frac{I\overrightarrow{dl} \sin \theta}{r^2} \) (1)
Where \( r \) is the distance from \( dl \) to P, and \( \theta \) is the angle between \( \overrightarrow{dl} \) and \( \overrightarrow{r} \).
The direction of the field \( d\overrightarrow{B} \) is perpendicular to the plane containing \( dl \) and \( r \), pointing into the page.
From the geometry in \( \triangle PAO \):
\( \tan(\pi - \theta) = \frac{a}{l} \)
\( \implies -\tan \theta = \frac{a}{l} \)
\( \implies l = \frac{-a}{\tan \theta} = -a \cot \theta \) (2)
Also from \( \triangle PAO \):
\( \sin \theta = \frac{a}{r} \)
\( \implies r = \frac{a}{\sin \theta} = a \operatorname{cosec} \theta \) (3)
Differentiating equation (2) with respect to \( \theta \):
\( dl = -a(-\operatorname{cosec}^2 \theta) d\theta = a \operatorname{cosec}^2 \theta d\theta \) (4)
Substitute (3) and (4) into equation (1):
\( dB = \frac{\mu_0 I}{4\pi} \frac{(a \operatorname{cosec}^2 \theta d\theta) \sin \theta}{(a \operatorname{cosec} \theta)^2} \)
\( dB = \frac{\mu_0 I}{4\pi} \frac{a \operatorname{cosec}^2 \theta \sin \theta}{a^2 \operatorname{cosec}^2 \theta} d\theta \)
\( dB = \frac{\mu_0 I}{4\pi a} \sin \theta d\theta \) (5)
To find the total magnetic field B, we integrate \( dB \) from \( \phi_1 \) to \( \phi_2 \), which are the angles defining the length of the wire from point P. For an infinitely long straight conductor, \( \phi_1 = 0 \) and \( \phi_2 = \pi \).
\( B = \int_0^\pi \frac{\mu_0 I}{4\pi a} \sin \theta d\theta \)
\( B = \frac{\mu_0 I}{4\pi a} [-\cos \theta]_0^\pi \)
\( B = \frac{\mu_0 I}{4\pi a} (-\cos \pi - (-\cos 0)) \)
\( B = \frac{\mu_0 I}{4\pi a} (-(-1) - (-1)) \)
\( B = \frac{\mu_0 I}{4\pi a} (1 + 1) \)
\( B = \frac{\mu_0 I}{4\pi a} (2) \)
\( B = \frac{\mu_0 I}{2\pi a} \)
The total magnetic field at point P due to an infinitely long straight conductor is \( B = \frac{\mu_0 I}{2\pi a} \). This expression shows that the magnetic field strength decreases as you move further away from the wire.
In simple words: For a very long straight wire carrying current, the magnetic field around it gets weaker as you move further away. The formula \( B = \frac{\mu_0 I}{2\pi a} \) gives us this field, where \( I \) is current and \( a \) is distance from the wire.

🎯 Exam Tip: Remember to use Biot-Savart's law for elemental length and then integrate over the entire length of the conductor, adjusting limits for an infinitely long wire.

 

Question 3. Obtain a relation for the magnetic field at a point along the axis of a circular coil carrying current.
Answer:Let's find the magnetic field at a point along the axis of a circular coil.
1. Consider a circular coil with radius R, carrying a current I.
2. Let P be a point on the axis of the coil at a distance z from its center 'O'.
3. Consider two small, diametrically opposite current elements \( d\overrightarrow{l} \) at points C and D on the coil.
4. Let \( \overrightarrow{r} \) be the vector joining the current element \( I d\overrightarrow{l} \) to point P. From the figure, we see that \( PC = PD = r = \sqrt{R^2 + z^2} \). The angles \( \angle CPO \) and \( \angle DPO \) are equal, let's call them \( \theta \).
5. According to Biot-Savart's law, the magnetic field \( d\overrightarrow{B} \) due to a current element is given by: \( d\overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{I d\overrightarrow{l} \times \hat{r}}{r^2} \)
6. Each \( d\overrightarrow{B} \) can be split into two components: \( dB \sin \theta \) (along the y-direction, perpendicular to the axis) and \( dB \cos \theta \) (along the z-direction, parallel to the axis).
7. Due to the diametrically opposite elements, the perpendicular components \( dB \sin \theta \) will cancel each other out. Only the components parallel to the axis, \( dB \cos \theta \), will add up to contribute to the total magnetic field.
From the figure, \( \cos \theta = \frac{R}{r} = \frac{R}{\sqrt{R^2 + z^2}} \).
8. Integrating the axial components over the entire loop: \( B = \int dB \cos \theta \)
\( B = \int \frac{\mu_0 I}{4\pi r^2} dl \cos \theta \)
Substitute \( \cos \theta \) and \( r \):
\( B = \int \frac{\mu_0 I}{4\pi (R^2 + z^2)} dl \frac{R}{\sqrt{R^2 + z^2}} \)
\( B = \frac{\mu_0 I R}{4\pi (R^2 + z^2)^{3/2}} \int dl \)
The integral \( \int dl \) is the total length of the circular coil, which is its circumference, \( 2\pi R \).
\( B = \frac{\mu_0 I R}{4\pi (R^2 + z^2)^{3/2}} (2\pi R) \)
\( B = \frac{\mu_0 I R^2}{2 (R^2 + z^2)^{3/2}} \)
If the coil has N turns, then the magnetic field is multiplied by N:
\( B = \frac{\mu_0 N I R^2}{2 (R^2 + z^2)^{3/2}} \)
This is the magnetic field at a point along the axis of a circular coil. If \( z \gg R \), then \( (R^2+z^2)^{3/2} \approx (z^2)^{3/2} = z^3 \).
So, \( B \approx \frac{\mu_0 N I R^2}{2 z^3} \). This formula helps us understand how the field changes with distance along the axis.
In simple words: The magnetic field along the center line of a circular coil carrying current depends on the current, the coil's size, and how far you are from it. The field gets weaker as you move further away.

🎯 Exam Tip: Remember that for a circular coil, the perpendicular components of \( d\overrightarrow{B} \) cancel out, and only the axial components contribute to the net magnetic field. Also, the term \( (R^2 + z^2)^{3/2} \) is crucial.

 

Question 4. Compute the torque experienced by a magnetic needle in a uniform magnetic field.
Answer:Let's calculate the torque on a magnetic needle (bar magnet) in a uniform magnetic field.
1. Consider a bar magnet NS of magnetic length \( 2l \), with pole strength \( q_m \). It is placed in a uniform magnetic field \( \overrightarrow{B} \).
2. The north pole (N) experiences a force \( \overrightarrow{F_N} = q_m \overrightarrow{B} \) in the direction of the field.
3. The south pole (S) experiences a force \( \overrightarrow{F_S} = -q_m \overrightarrow{B} \) in the opposite direction to the field.
4. The net force on the magnet is \( \overrightarrow{F} = \overrightarrow{F_N} + \overrightarrow{F_S} = q_m \overrightarrow{B} - q_m \overrightarrow{B} = 0 \). So, there is no translatory motion.
5. However, since the two forces are equal, opposite, and acting at different points, they form a couple. This couple creates a torque that tends to rotate the magnet.
6. Let the magnetic moment \( \overrightarrow{p_m} \) make an angle \( \theta \) with the magnetic field \( \overrightarrow{B} \). The torque \( \overrightarrow{\tau} \) is given by: \( \overrightarrow{\tau} = \overrightarrow{ON} \times \overrightarrow{F_N} + \overrightarrow{OS} \times \overrightarrow{F_S} \)
Where O is the center of the magnet. Using the right-hand corkscrew rule, the torque points into the page. The magnitude of the torque from each pole about the center O is \( l F \sin \theta = l (q_m B) \sin \theta \).
The total torque \( \tau = (l)(q_m B \sin \theta) + (l)(q_m B \sin \theta) \)
\( \tau = 2 l q_m B \sin \theta \)
Since magnetic dipole moment \( p_m = q_m (2l) \), we can write the torque as:
\( \tau = p_m B \sin \theta \)
In vector form, \( \overrightarrow{\tau} = \overrightarrow{p_m} \times \overrightarrow{B} \).
This torque makes the magnetic needle rotate until its magnetic moment aligns with the external magnetic field. This formula is important for understanding how compasses work.
In simple words: A magnet in a steady magnetic field feels a twisting force (torque) that tries to line it up with the field. This twisting force is strongest when the magnet is at an angle to the field.

🎯 Exam Tip: Remember that a magnetic dipole in a uniform magnetic field experiences a net torque, but no net force, causing it to align with the field. The formula \( \tau = p_m B \sin \theta \) is key.

 

Question 5. Calculate the magnetic field at a point on the axial line of a bar magnet.
Answer:Let's calculate the magnetic field at a point on the axial line of a bar magnet.
1. Consider a bar magnet NS with a magnetic length \( 2l \) and pole strength \( q_m \). N is the Northpole and S is the Southpole.
2. Let C be a point located along the axial line of the magnet at a distance \( r \) from its center O.
3. According to Coulomb’s law of magnetism, the force of repulsion (magnetic field) due to the North pole at C is: \( \overrightarrow{B_N} = \frac{\mu_0}{4\pi} \frac{q_m}{(r-l)^2} \hat{i} \) (1)
Here, \( (r-l) \) is the distance from the North pole to point C. The direction is outwards from N.
4. The force of attraction (magnetic field) due to the South pole at C is: \( \overrightarrow{B_S} = \frac{\mu_0}{4\pi} \frac{q_m}{(r+l)^2} (-\hat{i}) \) (2)
Here, \( (r+l) \) is the distance from the South pole to point C. The direction is towards S.
5. The net magnetic field \( \overrightarrow{B} \) at point C is the vector sum of these two fields:
\( \overrightarrow{B} = \overrightarrow{B_N} + \overrightarrow{B_S} \)
\( \overrightarrow{B} = \frac{\mu_0 q_m}{4\pi} \left[ \frac{1}{(r-l)^2} - \frac{1}{(r+l)^2} \right] \hat{i} \)
\( \overrightarrow{B} = \frac{\mu_0 q_m}{4\pi} \left[ \frac{(r+l)^2 - (r-l)^2}{(r-l)^2 (r+l)^2} \right] \hat{i} \)
\( \overrightarrow{B} = \frac{\mu_0 q_m}{4\pi} \left[ \frac{(r^2 + l^2 + 2rl) - (r^2 + l^2 - 2rl)}{(r^2 - l^2)^2} \right] \hat{i} \)
\( \overrightarrow{B} = \frac{\mu_0 q_m}{4\pi} \left[ \frac{4rl}{(r^2 - l^2)^2} \right] \hat{i} \)
\( \overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{2(q_m 2l)r}{(r^2 - l^2)^2} \hat{i} \)
Since magnetic dipole moment \( p_m = q_m (2l) \), we can write:
\( \overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{2 p_m r}{(r^2 - l^2)^2} \hat{i} \)
For a short bar magnet, \( r \gg l \), so \( (r^2 - l^2)^2 \approx (r^2)^2 = r^4 \).
Therefore, for a short bar magnet, the magnetic field on the axial line is:
\( \overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{2 p_m r}{r^4} \hat{i} \)
\( \overrightarrow{B} = \frac{\mu_0}{4\pi} \frac{2 p_m}{r^3} \hat{i} \)
This equation shows that the magnetic field decreases rapidly with distance from a short bar magnet along its axial line. This relationship helps understand how magnets interact.
In simple words: To find the magnetic field straight out from the ends of a bar magnet, you add the fields from its north and south poles. For a short magnet, this field gets much weaker the further you go.

🎯 Exam Tip: Remember to apply Coulomb's law for magnetism for each pole and then sum them vectorially. For a short magnet, the approximation \( r \gg l \) simplifies the final expression to \( B \propto \frac{1}{r^3} \).

 

Question 6. Obtain the magnetic field at a point on the equatorial line of a bar magnet.
Answer: To find the magnetic field along the equatorial line of a bar magnet, imagine a bar magnet labeled N (North pole) and S (South pole). Its pole strength is \( q_m \) and its total length is \( 2l \). Consider a point C along the equatorial line, which is an imaginary line perpendicular to the magnet's axis and passing through its center. The distance from the center of the magnet to point C is \( r \). Using Coulomb's law of magnetism, we calculate the forces of repulsion and attraction from each pole on a unit north pole placed at point C. These forces, \( F_N \) from the North pole and \( F_S \) from the South pole, have equal magnitudes but act in different directions. The magnetic field at any point on the equatorial line is always perpendicular to the magnetic axis.

This diagram shows a bar magnet NS with its geometrical center O. Point C is on the equatorial line at a distance r from O. The forces \( \vec{F}_N \) and \( \vec{F}_S \) due to the North and South poles, respectively, are shown, along with their components.
The force of repulsion from the North pole at C is:
\( \overrightarrow{F_N} = -F_N \cos \theta \hat{i} + F_N \sin \theta \hat{j} \)
Where,
\( F_N = \frac{\mu_0}{4 \pi} \frac{q_m}{r'^2} \) …………(1)
Here, \( r' \) is the distance from the north pole to point C.
The force of attraction from the South pole at C is:
\( \overrightarrow{F_S} = -F_S \cos \theta \hat{i} - F_S \sin \theta \hat{j} \)
Where,
\( F_S = \frac{\mu_0}{4 \pi} \frac{q_m}{r'^2} \) ……………(2)
The net force at point C is the sum of these two forces:
\( \overrightarrow{F} = \overrightarrow{F_N} + \overrightarrow{F_S} \)
Combining the components, the \( \hat{j} \) components cancel out, and the \( \hat{i} \) components add up.
\( \overrightarrow{B} = - (F_N + F_S) \cos \theta \hat{i} \)
Since \( F_N = F_S \), we have:
\( \overrightarrow{B} = -\frac{2 \mu_0}{4 \pi} \frac{q_m}{r'^2} \cos \theta \hat{i} \)
From the geometry, \( r'^2 = r^2 + l^2 \) and \( \cos \theta = \frac{l}{r'} = \frac{l}{\sqrt{r^2+l^2}} \)
So,
\( \overrightarrow{B} = -\frac{2 \mu_0}{4 \pi} \frac{q_m}{(r^2+l^2)} \frac{l}{\sqrt{r^2+l^2}} \hat{i} \)
\( \overrightarrow{B} = -\frac{\mu_0}{4 \pi} \frac{2 q_m l}{(r^2+l^2)^{3/2}} \hat{i} \)
We know that the magnetic dipole moment \( p_m = q_m (2l) \).
Therefore,
\( \overrightarrow{B} = -\frac{\mu_0}{4 \pi} \frac{p_m}{(r^2+l^2)^{3/2}} \hat{i} \)
If the magnet is very short compared to the distance C (i.e., \( r >> l \)), then \( l^2 \) can be ignored.
\( (r^2+l^2)^{3/2} \approx (r^2)^{3/2} = r^3 \)
So, for a short magnet, the magnetic field is:
\( \overrightarrow{B} = -\frac{\mu_0}{4 \pi} \frac{p_m}{r^3} \hat{i} \)
This expression shows that the magnetic field on the equatorial line is opposite to the direction of the magnetic moment and decreases rapidly with distance.
In simple words: We calculate the pull and push from both ends of the magnet at a point on its middle line. The sideways forces cancel out, and only the forces pointing directly along the line remain. This gives us a formula for the magnetic field strength, which becomes weaker the further away you go.

🎯 Exam Tip: Remember to clearly define all variables (like \( q_m \), \( 2l \), \( r \), \( r' \), and \( p_m \)) and show how the components of the forces are resolved. Pay attention to the negative sign in the final expression, indicating the direction of the field relative to the magnetic moment.

 

Question 7. Find the magnetic field due to a long straight conductor using Ampere’s circuital law.
Answer: To find the magnetic field from a long straight wire, we use Ampere's circuital law.
1. Imagine a very long, straight wire (NM) carrying an electric current \( I \).
2. The wire is shaped like a cylinder and has a uniform current distribution.
3. To find the magnetic field, we draw a circular path around the wire, called an Amperian loop, at a distance \( r \) from the wire's center. This loop is symmetrical.

This diagram shows a straight current-carrying conductor in the center, with a circular Amperian loop of radius r around it, along with the magnetic field B and elemental length dl vectors.

4. According to Ampere’s circuital law, the line integral of the magnetic field \( \overrightarrow{B} \) around any closed loop is equal to \( \mu_0 \) times the net current \( I_{enclosed} \) passing through the loop.
\( \oint \overrightarrow{B} \cdot \overrightarrow{dl} = \mu_0 I_{enclosed} \)
5. For our circular Amperian loop, the magnetic field \( \overrightarrow{B} \) is uniform (constant in magnitude) at every point on the loop, and its direction is always parallel to the elemental length vector \( \overrightarrow{dl} \). This means the angle between \( \overrightarrow{B} \) and \( \overrightarrow{dl} \) is \( 0^\circ \), so \( \cos 0^\circ = 1 \).
6. The integral becomes:
\( B \oint dl = \mu_0 I \)
7. The integral \( \oint dl \) is simply the total length of the Amperian loop, which is its circumference, \( 2 \pi r \).
\( B (2 \pi r) = \mu_0 I \)
8. Solving for \( B \), we get the magnetic field strength:
\( B = \frac{\mu_0 I}{2 \pi r} \)
In vector form, considering \( \hat{n} \) as the unit vector tangent to the loop, the magnetic field is:
\( \overrightarrow{B} = \frac{\mu_0 I}{2 \pi r} \hat{n} \)
This shows that the magnetic field strength decreases as you move further away from the wire.
In simple words: We draw a circle around the wire. The magnetic field is the same all around this circle. By using a rule called Ampere's law, we can easily find out how strong the magnetic field is at any distance from the wire. The field gets weaker as you move further from the wire.

🎯 Exam Tip: When applying Ampere's law, always choose an Amperian loop where the magnetic field is either constant and parallel to the loop's path, or perpendicular to it (so the dot product is zero), to simplify the calculation.

 

Question 8. Discuss the working of a cyclotron in detail.
Answer: A cyclotron is a special device used to speed up charged particles, like protons or ions, to very high kinetic energies. It's also known as a high-energy accelerator.

**Principle:**
When a charged particle moves into a magnetic field at a right angle, it experiences a magnetic force called the Lorentz force. This force makes the particle move in a circle, but it doesn't change the particle's speed.

**Construction:**
1. The main parts are two D-shaped metal containers, called "Dees." Particles are allowed to move between these Dees.
2. A strong, uniform magnetic field is created by an electromagnet, which is perpendicular to the plane of the Dees.
3. A source 'S' for charged particles is placed in the gap between the two Dees.
4. The Dees are connected to a high-frequency alternating voltage, which changes its polarity very quickly.

**Working:**
1. First, a positively charged ion is released from the source in the center.
2. One Dee (let's say Dee-1) is negative, attracting the ion and pulling it inside.
3. Inside the Dee, the magnetic field makes the ion move in a semi-circular path. The centripetal force needed for this circular motion is provided by the Lorentz force:
\( \frac{mv^2}{r} = Bqv \)
\( r = \frac{mv}{Bq} \)
This shows that the radius of the path is directly proportional to the particle's speed.
4. When the ion reaches the gap again, the polarity of the Dees reverses. Dee-1 becomes positive, and Dee-2 becomes negative. This gives the ion another push, accelerating it towards Dee-2.
5. The ion's speed increases, so its path inside Dee-2 is a larger semi-circle. This process repeats: the ion gets accelerated across the gap, its speed increases, and it moves in larger semi-circles.

This diagram illustrates the basic components of a cyclotron, showing the electromagnet poles (N and S), the D-shaped Dees, the ion source, and the spiral path of accelerated particles.
6. As the radius of the circular path grows, the particle's velocity also increases, making it follow a spiral path.
7. Once the particles reach the maximum radius and desired energy, a deflector plate guides them out to hit a target.

**Resonance condition:**
For the cyclotron to work, the frequency at which the positive ions complete a semi-circle must match the frequency of the alternating voltage applied to the Dees. This is called the resonance condition.
\( f_{osc} = \frac{qB}{2 \pi m} \)
The time period of oscillation is \( T = \frac{2 \pi m}{qB} \).
The kinetic energy of the charged particle is \( KE = \frac{1}{2} mv^2 = \frac{q^2 B^2 r^2}{2m} \).

**Limitations:**
1. The speed of the ions that can be accelerated is limited because as particles get faster, their mass effectively increases (relativistic effect), disrupting the resonance.
2. Electrons cannot be accelerated in a cyclotron because their mass is very small and changes significantly even at relatively low speeds, making it difficult to maintain the resonance condition.
3. Uncharged particles also cannot be accelerated since they do not experience the Lorentz force. The cyclotron uses electric and magnetic fields to accelerate charged particles, making it ideal for experiments in nuclear physics.
In simple words: A cyclotron makes tiny charged particles go faster and faster in a spiral path. It uses strong magnets to keep them moving in circles and electric pushes to speed them up each time they cross a gap. But it can't speed up electrons or uncharged particles because of how it works.

🎯 Exam Tip: When explaining the cyclotron, highlight the role of both the magnetic field (for circular path) and the alternating electric field (for acceleration across the gap). Remember the resonance condition and the key reasons why electrons cannot be accelerated.

 

Question 9. What is tangent law? Discuss in detail.
Answer: **Tangent Law:**
When a magnetic needle or a small magnet is freely hung and placed in two magnetic fields that are perpendicular to each other, it will settle down in the direction of the combined (resultant) magnetic field.
If \( B_H \) is the horizontal component of Earth’s magnetic field and \( B \) is the magnetic field produced by the tangent galvanometer coil, and the needle deflects by an angle \( \theta \), then according to the tangent law:
\( B = B_H \tan \theta \)

**Construction of a Tangent Galvanometer:**
1. It has a circular coil made of wire, wrapped around a frame of brass or wood.
2. The whole device sits on a circular base that can be leveled using leveling screws. This ensures the coil is perfectly vertical.
3. A small compass box is placed at the center of the coil.
4. Inside the compass box, there's a small, pivoted magnet with an aluminum pointer attached to it.
5. The circular scale around the compass is divided into four sections (quadrants), each marked from 0 to 90 degrees.

**Theory:**
1. When the tangent galvanometer is set up correctly, the plane of its coil is aligned with Earth's magnetic meridian. This means the magnetic field produced by the coil will be perpendicular to Earth's horizontal magnetic field.
2. The compass needle, initially aligned with Earth's horizontal field \( B_H \), will deflect when a current passes through the coil, creating its own magnetic field \( B \).

This diagram shows a compass needle at the center of a circular coil. \( B_H \) represents the horizontal component of Earth's magnetic field, and \( B \) is the field produced by the coil. The needle deflects by an angle \( \theta \).
3. The magnetic induction \( B \) at the center of the coil due to the current \( I \) flowing through it is given by:
\( B = \frac{\mu_0 NI}{2R} \) …………(2)
Where \( \mu_0 \) is the permeability of free space, \( N \) is the number of turns in the coil, and \( R \) is the radius of the coil.
4. At equilibrium, the tangent of the deflection angle \( \theta \) is the ratio of the magnetic field due to the coil to Earth's horizontal magnetic field:
\( \tan \theta = \frac{B}{B_H} \)
5. Substituting the expression for \( B \):
\( \frac{\mu_0 NI}{2R} = B_H \tan \theta \)
6. From this, the current \( I \) can be calculated as:
\( I = \frac{2RB_H}{\mu_0 N} \tan \theta \)
This equation shows that the current is proportional to \( \tan \theta \). The tangent law is fundamental for tangent galvanometers, which measure electric currents based on magnetic field deflections.
In simple words: The tangent law helps us measure electric current. When a compass needle is put inside a special coil and a current flows, the needle moves. By looking at how much it moves, we can find the current using a simple math rule involving the "tangent" of the angle.

🎯 Exam Tip: Ensure you clearly state the condition for tangent law (two perpendicular magnetic fields) and derive the formula \( B = B_H \tan \theta \). Also, remember the factors affecting the galvanometer's sensitivity (like number of turns and coil radius).

 

Question 10. Derive the expression for the torque on a current-carrying coil in a magnetic field.
Answer: Let's find out how a current-carrying coil spins (experiences torque) when placed in a uniform magnetic field.
1. Imagine a rectangular loop named PQRS, carrying an electric current \( I \).
2. This loop is placed inside a uniform magnetic field \( B \). Let the length of the loop be \( a \) and its breadth be \( b \).
3. The unit vector \( \hat{n} \), which is perpendicular to the plane of the loop, makes an angle \( \theta \) with the direction of the magnetic field.

**Forces on the sides of the coil:**
Each side of the coil experiences a magnetic force due to the current flowing in the magnetic field.
1. **Arm PQ:** The magnetic force acting on this arm is \( F_{PQ} = IaB \sin (\pi/2) = IaB \). Its direction is outwards, perpendicular to the plane of the loop.
2. **Arm QR:** The force is \( F_{QR} = IbB \sin (\pi/2 - \theta) = IbB \cos \theta \).
3. **Arm RS:** The force is \( F_{RS} = IaB \sin (\pi/2) = IaB \). Its direction is inwards, perpendicular to the plane of the loop.
4. **Arm SP:** The force is \( F_{SP} = IbB \sin (\pi/2 - \theta) = IbB \cos \theta \).

This diagram shows a rectangular coil PQRS placed in a uniform magnetic field. The forces on each side of the coil are indicated. The side view shows the angle \( \theta \) between the normal to the coil and the magnetic field B.

**Net force and Torque:**
5. The forces \( F_{QR} \) and \( F_{SP} \) are equal in magnitude, opposite in direction, and act along the same line (collinear). Therefore, they cancel each other out, and produce no net torque.
6. The forces \( F_{PQ} \) and \( F_{RS} \) are also equal in magnitude and opposite in direction, but they are not collinear. They form a "couple" and produce a torque that rotates the coil.
7. The total torque \( \tau \) acting on the loop about its axis AB is the sum of the torques from arms PQ and RS.
\( \tau = (\frac{b}{2} \sin \theta) F_{PQ} + (\frac{b}{2} \sin \theta) F_{RS} \)
Since \( F_{PQ} = F_{RS} = IaB \):
\( \tau = (\frac{b}{2} \sin \theta) IaB + (\frac{b}{2} \sin \theta) IaB \)
\( \tau = IaB (b \sin \theta) \)
Since the area of the coil \( A = a \times b \), we can write the torque as:
\( \tau = IAB \sin \theta \)
In vector form, the torque is:
\( \overrightarrow{\tau} = (\overrightarrow{IA}) \times \overrightarrow{B} \)
8. This equation can also be written using the magnetic dipole moment \( \overrightarrow{p_m} \), where \( \overrightarrow{p_m} = \overrightarrow{IA} \). So,
\( \overrightarrow{\tau} = \overrightarrow{p_m} \times \overrightarrow{B} \)
9. If the rectangular loop has \( N \) turns, the total torque is:
\( \tau = NIAB \sin \theta \)

**Special cases:**
* If \( \theta = 90^\circ \) (the plane of the loop is parallel to the magnetic field), then \( \sin 90^\circ = 1 \), and the torque is maximum: \( \tau_{max} = NIAB \).
* If \( \theta = 0^\circ \) or \( 180^\circ \) (the plane of the loop is perpendicular to the magnetic field), then \( \sin 0^\circ = 0 \), and the torque on the current loop is zero. This torque is crucial for understanding how electric motors and galvanometers operate, converting electrical energy into mechanical rotation.
In simple words: When you put a loop of wire with current in a magnetic field, the field pushes on the sides of the loop. These pushes create a twisting force, called torque, that makes the loop want to spin. The amount of spin depends on the current, the area of the loop, the magnetic field strength, and the angle the loop makes with the field.

🎯 Exam Tip: Clearly identify the forces on each side of the loop, explaining which ones cancel and which ones contribute to the torque. Show the derivation of \( \tau = NIAB \sin \theta \) and discuss the maximum and minimum torque conditions for full marks.

 

Question 11. Discuss the conversion of a galvanometer into an ammeter and also a voltmeter.
Answer: A galvanometer is a device that can detect and measure small electric currents. It can be modified to measure larger currents (as an ammeter) or voltages (as a voltmeter).

**Galvanometer to an Ammeter:**
1. An ammeter needs to have very low resistance so that it does not significantly change the current it is measuring in a circuit.
2. To convert a galvanometer into an ammeter, a very small resistance, called a "shunt resistance" (\( S \)), is connected in parallel with the galvanometer.
3. The shunt resistance allows most of the current to bypass the galvanometer, while a small, measurable current flows through the galvanometer itself.

This diagram illustrates a galvanometer (G) connected in parallel with a shunt resistor (S) to form an ammeter. The total current (I) splits, with \( I_g \) flowing through the galvanometer and \( I_s \) through the shunt.

Let \( I \) be the total current in the circuit, \( I_g \) be the current through the galvanometer, and \( I_s \) be the current through the shunt.
\( I_s = I - I_g \)
Since the shunt and galvanometer are in parallel, the voltage across them is the same:
\( I_g R_g = I_s S \)
\( I_g R_g = (I - I_g) S \)
The required shunt resistance is:
\( S = \frac{I_g R_g}{I - I_g} \) ………….(1)
The effective resistance of the ammeter \( R_a \) is very low, ideally zero for an ideal ammeter.

**Galvanometer to a Voltmeter:**
1. A voltmeter needs to have very high resistance so that it draws negligible current from the circuit and measures the potential difference accurately.
2. To convert a galvanometer into a voltmeter, a very large resistance (\( R_h \)) is connected in series with the galvanometer.
3. This high resistance limits the current flowing through the galvanometer and the circuit, allowing the voltmeter to measure the voltage across a component without affecting the circuit significantly.
This diagram depicts a galvanometer (G) connected in series with a high resistance \( R_h \) to form a voltmeter. The current \( I_g \) flows through both.

Let \( V \) be the voltage to be measured. The total resistance in the series circuit is \( R_g + R_h \).
The current flowing through the galvanometer is:
\( I_g = \frac{V}{R_g + R_h} \)
The required high resistance is:
\( R_h = \frac{V}{I_g} - R_g \)
An ideal voltmeter has infinite resistance. The conversion process is vital in electronics for creating versatile measurement tools from a basic galvanometer.
In simple words: A normal galvanometer can be changed into two different tools. To make it an ammeter (to measure current), you add a very small resistor next to it. To make it a voltmeter (to measure voltage), you add a very large resistor in a line with it.

🎯 Exam Tip: Clearly explain why ammeters need low resistance (parallel shunt) and voltmeters need high resistance (series resistor). Remember the formulas for calculating the required shunt and series resistances.

 

Question 12. Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law.
Answer: We can find the magnetic field inside and outside a long solenoid using Ampere's circuital law.
1. Consider a very long solenoid with length \( L \) and \( N \) turns of wire. The solenoid's diameter is much smaller than its length.

**Magnetic field inside the solenoid:**
2. To find the magnetic field inside, we draw a rectangular Amperian loop (abcd) inside the solenoid, with side 'h' parallel to the solenoid's axis and sides 'b' and 'd' perpendicular to the axis.

This diagram illustrates a long solenoid carrying current. An Amperian loop 'abcd' is drawn inside, with side 'h' parallel to the solenoid axis. The magnetic field is strong and uniform inside, and negligible outside.

3. Ampere’s circuital law states: \( \oint \overrightarrow{B} \cdot \overrightarrow{dl} = \mu_0 I_{enclosed} \).
4. We break the integral into four parts, one for each side of the rectangle:
\( \oint_C \overrightarrow{B} \cdot \overrightarrow{dl} = \int_a^b \overrightarrow{B} \cdot \overrightarrow{dl} + \int_b^c \overrightarrow{B} \cdot \overrightarrow{dl} + \int_c^d \overrightarrow{B} \cdot \overrightarrow{dl} + \int_d^a \overrightarrow{B} \cdot \overrightarrow{dl} \)
5. For the segments bc and da, the magnetic field is perpendicular to \( \overrightarrow{dl} \), so \( \cos 90^\circ = 0 \). Thus, \( \int_b^c \overrightarrow{B} \cdot \overrightarrow{dl} = 0 \) and \( \int_d^a \overrightarrow{B} \cdot \overrightarrow{dl} = 0 \).
6. For the segment cd, which is outside the solenoid (or if inside, it is assumed to be very small, close to zero), the magnetic field is approximately zero. So, \( \int_c^d \overrightarrow{B} \cdot \overrightarrow{dl} = 0 \).
7. For the segment ab (of length \( h \)) inside the solenoid, \( \overrightarrow{B} \) is parallel to \( \overrightarrow{dl} \), and \( B \) is uniform. So, \( \int_a^b \overrightarrow{B} \cdot \overrightarrow{dl} = Bh \).
8. The total current enclosed by the loop is \( I_{enclosed} = nIh \), where \( n \) is the number of turns per unit length ( \( n = N/L \) ).
9. Applying Ampere’s law:
\( Bh = \mu_0 nIh \)
\( B = \mu_0 nI \)
This shows the magnetic field inside a long solenoid is uniform and depends on the current and number of turns per unit length.

**Magnetic field outside the solenoid:**
1. To find the magnetic field outside, consider an Amperian loop enclosing the entire solenoid.
2. For an ideal long solenoid, the magnetic field lines are concentrated almost entirely inside the solenoid and are negligible outside.
3. Therefore, outside the solenoid, the magnetic field \( B \) is considered to be approximately zero. Magnetic fields are crucial for many technologies, from electromagnets to medical imaging.
In simple words: To find the magnetic field inside a long coil (solenoid), we use a special rule called Ampere's law. We draw an imaginary box inside the coil, and it helps us find that the magnetic field is strong and even. Outside the coil, the magnetic field is practically zero.

🎯 Exam Tip: When deriving the magnetic field for a solenoid, clearly define your Amperian loop and explain why the contributions from each segment of the loop are what they are. Emphasize that the field outside an ideal long solenoid is zero.

 

Question 13. Derive the expression for the force between two parallel, current-carrying conductors.
Answer: To find the force between two parallel current-carrying conductors, we consider two long, parallel wires, XY and PQ. These wires carry steady currents \( I_1 \) and \( I_2 \) respectively, in the same direction, and are separated by a distance \( d \) in the air. The magnetic field \( B_1 \) produced by current \( I_1 \) at any point on the conductor PQ can be written as:
\( B_1 = \frac{\mu_0 I_1}{2\pi d} \)
The direction of this magnetic field \( B_1 \) is perpendicular to the plane containing the conductors and points inwards. Now, the force experienced by a small elemental length \( dl \) of the conductor PQ due to this magnetic field \( B_1 \) is given by Lorentz force:
\( d\vec{F} = I_2 (\vec{dl} \times \vec{B}_1) \)
Since the current \( I_2 \) in PQ is perpendicular to \( B_1 \), the magnitude of the force \( dF \) on length \( dl \) is:
\( dF = I_2 dl B_1 \sin(90^\circ) = I_2 dl B_1 \)
Substitute the expression for \( B_1 \):
\( dF = I_2 dl \left( \frac{\mu_0 I_1}{2\pi d} \right) = \frac{\mu_0 I_1 I_2}{2\pi d} dl \)
The force per unit length \( \frac{F}{L} \) on conductor PQ is:
\( \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \)
Using Fleming's left-hand rule, we find that the force on PQ due to the magnetic field of XY is directed towards XY. This means the two wires attract each other when currents flow in the same direction. If the currents flow in opposite directions, the force will be repulsive, pushing the wires apart. This fundamental interaction explains how electric currents can exert forces on each other without direct contact.
In simple words: When two wires carry electric current in the same direction and are placed next to each other, they pull towards each other. The amount of this pulling force depends on how much current is flowing in each wire and how far apart they are. If the currents flow in opposite directions, they push each other away.

🎯 Exam Tip: Remember that parallel currents attract and anti-parallel currents repel. Always show the clear steps for \(B_1\) and then for \(F\) per unit length.

 

Question 14. Give an account of magnetic Lorentz force.
Answer: The magnetic Lorentz force is the force that a magnetic field exerts on a charged particle moving through it. When an electric charge \( q \) moves with a velocity \( \vec{v} \) in a magnetic field \( \vec{B} \), it experiences a force \( \vec{F}_m \). This force is given by the cross product:
\( \vec{F}_m = q (\vec{v} \times \vec{B}) \)
The magnitude of this force is \( F_m = qvB \sin\theta \), where \( \theta \) is the angle between the velocity vector \( \vec{v} \) and the magnetic field vector \( \vec{B} \). This force is always perpendicular to both the velocity of the particle and the magnetic field direction. Because the force is perpendicular to velocity, the magnetic Lorentz force does no work on the charged particle, meaning it only changes the direction of the particle's motion, not its speed or kinetic energy. The direction of the force depends on the sign of the charge: for a positive charge, it follows the right-hand rule (from \( \vec{v} \) to \( \vec{B} \)), while for a negative charge, the force is in the opposite direction. If the charge moves parallel or anti-parallel to the magnetic field (\( \theta = 0^\circ \) or \( 180^\circ \)), the magnetic Lorentz force is zero. This force is crucial for understanding how particles behave in accelerators and how electric motors work.
In simple words: The magnetic Lorentz force is a push or pull on a moving electric charge when it is inside a magnetic field. This force makes the charge change its direction but does not make it go faster or slower. Its direction can be found using the right-hand rule.

🎯 Exam Tip: Clearly state the vector form \( \vec{F}_m = q (\vec{v} \times \vec{B}) \) and its magnitude \( F_m = qvB \sin\theta \). Emphasize that it does no work and only changes direction.

 

Question 15. Compare the properties of soft and hard ferromagnetic materials.
Answer:

In simple words: Soft ferromagnetic materials are easy to magnetize and demagnetize, like the cores in transformers. Hard ferromagnetic materials are difficult to magnetize but stay magnetized for a long time, making them good for permanent magnets.

🎯 Exam Tip: When comparing, ensure you list properties like retentivity, coercivity, and hysteresis loss, as these are key distinguishing features for these materials.

 

Question 16. Derive an expression for the force acting on a current-carrying conductor placed in a magnetic field.
Answer: To derive the expression for the force on a current-carrying conductor in a magnetic field, let's consider a straight wire of length \( l \) and cross-sectional area \( A \). The wire carries a current \( I \). Imagine the wire placed in a uniform magnetic field \( \vec{B} \). Inside the conductor, there are free electrons, each with charge \( -e \), moving with an average drift velocity \( \vec{v}_d \) in a direction opposite to the conventional current. Let \( n \) be the number of free electrons per unit volume.
The total number of free electrons in the small elemental length \( dl \) of the wire is \( N = n(A dl) \).
The Lorentz force experienced by a single electron is \( \vec{f} = -e(\vec{v}_d \times \vec{B}) \).
The total force \( d\vec{F} \) on the elemental length \( dl \) of the conductor is the sum of the forces on all electrons within that length:
\( d\vec{F} = N \vec{f} = (n A dl) [-e(\vec{v}_d \times \vec{B})] \)
Rearranging the terms, we get:
\( d\vec{F} = (-e n A \vec{v}_d) \times \vec{B} dl \)
We know that the current \( I \) is related to the drift velocity by \( I = ne A v_d \). The direction of \( I dl \) is the direction of \( \vec{v}_d \) for positive charges, so for electrons, \( I dl \) is in the opposite direction to \( \vec{v}_d \). Therefore, we can write \( \vec{I} \vec{dl} = -neA \vec{v}_d dl \).
Substituting this into the force equation:
\( d\vec{F} = (I \vec{dl}) \times \vec{B} \)
Integrating over the entire length \( l \) of the conductor, the total force \( \vec{F} \) on the current-carrying conductor is:
\( \vec{F} = I (\vec{l} \times \vec{B}) \)
The magnitude of this force is \( F = BIl \sin\theta \), where \( \theta \) is the angle between the length vector \( \vec{l} \) (in the direction of current) and the magnetic field \( \vec{B} \). This force is responsible for the operation of electric motors and other electromagnetic devices, converting electrical energy into mechanical motion. The force is zero if the conductor is parallel to the magnetic field (\( \theta = 0^\circ \)) and maximum if it is perpendicular (\( \theta = 90^\circ \)).
In simple words: When a wire carrying electric current is placed inside a magnetic field, it feels a push or pull. The strength of this force depends on how much current is in the wire, its length, the strength of the magnetic field, and the angle between the wire and the field. This force is what makes electric motors spin.

🎯 Exam Tip: Always show the steps clearly, starting from the force on individual electrons and linking it to the current in the conductor. Ensure vector notation is used for the force, current element, and magnetic field.

IV. Numerical problems:

 

Question 1. A bar magnet having a magnetic moment \( p_m \) is cut into four pieces i.e., first cut in two pieces along the axis of the magnet and each piece is further cut into two pieces. Compute the magnetic moment of each piece.
Answer: Let the original bar magnet have a magnetic moment \( p_m \), which is defined as the product of pole strength \( M \) and magnetic length \( l \), so \( p_m = M \times l \).
When the magnet is first cut into two pieces along its axis, it means the magnet is cut longitudinally. This halves the pole strength of each new piece, while the magnetic length remains the same. So, each of these two pieces will have a pole strength of \( M' = \frac{M}{2} \) and a magnetic length of \( l' = l \).
The magnetic moment of each of these two pieces would be \( p_m' = M' \times l' = \frac{M}{2} \times l = \frac{p_m}{2} \).
Next, each of these two pieces is further cut into two pieces. Since the cutting is not specified as along the axis or perpendicular, let's assume it means cutting each half-pole-strength piece into two equal lengths. This would halve the magnetic length again. So, each resulting piece would have a magnetic length of \( l'' = \frac{l}{2} \), and the pole strength would still be \( M' = \frac{M}{2} \).
The magnetic moment of each of these final four pieces would be \( p_m'' = M'' \times l'' = \frac{M}{2} \times \frac{l}{2} = \frac{M \times l}{4} = \frac{p_m}{4} \).
Therefore, each of the four final pieces will have a magnetic moment of \( \frac{p_m}{4} \). Magnetic moment is a vector quantity that represents the strength and orientation of a magnetic field source. This shows how dividing a magnet also divides its overall magnetic strength.
In simple words: When a magnet is cut first lengthwise into two pieces, and then each of those pieces is cut in half again, the original magnet is split into four smaller magnets. Each of these smaller magnets will have a magnetic strength that is one-fourth of the original magnet's strength.

🎯 Exam Tip: Clearly state how both pole strength and magnetic length change when a magnet is cut along its axis versus perpendicular to its axis. Here, assume a longitudinal cut first, then a transverse cut.

 

Question 2. A conductor of linear mass density 0.2 g m\(^{-1}\) suspended by two flexible wire as shown in figure. Suppose the tension in the supporting wires is zero when it is kept inside the magnetic field of 1 T whose direction is into the page. Compute the current inside the conductor and also the direction for the current.
Answer:To have zero tension in the supporting wires, the upward magnetic force per unit length on the conductor must exactly balance its downward weight per unit length.
Given:
Linear mass density \( \frac{m}{L} = 0.2 \text{ g m}^{-1} = 0.2 \times 10^{-3} \text{ kg m}^{-1} \)
Magnetic field strength \( B = 1 \text{ T} \)
Acceleration due to gravity \( g = 10 \text{ m s}^{-2} \)
The force per unit length due to gravity (weight per unit length) is:
\( \frac{F_g}{L} = \frac{m}{L} g = (0.2 \times 10^{-3} \text{ kg m}^{-1}) \times (10 \text{ m s}^{-2}) = 2 \times 10^{-3} \text{ N m}^{-1} \)
The magnetic force per unit length on a current-carrying conductor in a magnetic field is:
\( \frac{F_m}{L} = BI \sin\theta \)
For the magnetic force to be upwards and balance the weight, the angle \( \theta \) between the current and the magnetic field must be \( 90^\circ \), so \( \sin\theta = 1 \).
Therefore, \( \frac{F_m}{L} = BI \)
Equating the magnetic force per unit length to the weight per unit length for zero tension:
\( BI = \frac{m}{L} g \)
\( I = \frac{(m/L) g}{B} = \frac{2 \times 10^{-3} \text{ N m}^{-1}}{1 \text{ T}} = 2 \times 10^{-3} \text{ A} = 2 \text{ mA} \)
The current required is \( 2 \text{ mA} \).
To determine the direction of the current, we use Fleming's left-hand rule. The magnetic field is directed into the page (represented by crosses \( \times \)). The force is upwards. If you point your forefinger into the page and your thumb upwards, your middle finger will point to the right. Thus, the current must flow from left to right. This setup elegantly demonstrates the principle of magnetic levitation.
In simple words: To make the wires float without any tension, the upward push from the magnetic field must be equal to the wire's weight. We calculated that a current of 2 milliamperes is needed. For the wire to be pushed upwards when the magnetic field goes into the page, the current must flow from left to right.

🎯 Exam Tip: Always draw a simple diagram to visualize the forces and field directions. Remember to convert units to SI (grams to kilograms) before calculation to avoid errors.

 

Question 3. A circular coil with a cross-sectional area of 0.1 cm\(^2\) is kept in a uniform magnetic field of strength 0.2 T. If the current passing in the coil is 3A and the plane of the loop is perpendicular to the direction of magnetic field. Calculate
a) total torque on the coil
b) total force on the coil
c) average force on each electron in the coil due to the magnetic field. The free electron density for the material of the wire is 10\(^{28}\) m\(^{-3}\).
Answer: Given:
Number of turns \( N = 1 \)
Area \( A = 0.1 \text{ cm}^2 = 0.1 \times 10^{-4} \text{ m}^2 \)
Magnetic field \( B = 0.2 \text{ T} \)
Current \( I = 3 \text{ A} \)
Electron density \( n = 10^{28} \text{ m}^{-3} \)
The plane of the loop is perpendicular to the magnetic field. This means the normal to the plane of the coil is parallel to the magnetic field. So, the angle \( \theta \) between the magnetic dipole moment vector (normal to the coil plane) and the magnetic field vector is \( 0^\circ \).

a) Total torque on the coil:
The torque \( \tau \) on a current loop in a magnetic field is given by \( \tau = NIAB \sin\theta \).
Since \( \theta = 0^\circ \), \( \sin\theta = 0 \).
\( \tau = (1) \times (3 \text{ A}) \times (0.1 \times 10^{-4} \text{ m}^2) \times (0.2 \text{ T}) \times \sin(0^\circ) = 0 \)
Therefore, the total torque on the coil is \( 0 \text{ Nm} \). When the coil's normal aligns with the field, it is in stable equilibrium.

b) Total force on the coil:
For a closed current loop placed in a uniform magnetic field, the net magnetic force on the entire loop is always zero.
Therefore, the total force on the coil is \( 0 \text{ N} \).

c) Average force on each electron in the coil:
The magnetic force on a single electron is given by the Lorentz force formula \( F_e = |q| v_d B \sin\phi \), where \( q \) is the charge of the electron (\( -e = -1.6 \times 10^{-19} \text{ C} \)), \( v_d \) is the drift velocity, and \( \phi \) is the angle between \( \vec{v}_d \) and \( \vec{B} \). Since the plane of the coil is perpendicular to the magnetic field, the electrons' drift velocity (which is along the wire) will always be perpendicular to the magnetic field. So, \( \phi = 90^\circ \), and \( \sin\phi = 1 \).
We know the current \( I = neAv_d \). We can find the drift velocity \( v_d \):
\( v_d = \frac{I}{neA} = \frac{3 \text{ A}}{(10^{28} \text{ m}^{-3}) \times (1.6 \times 10^{-19} \text{ C}) \times (0.1 \times 10^{-4} \text{ m}^2)} \)
\( v_d = \frac{3}{1.6 \times 10^5} \text{ m/s} = 1.875 \times 10^{-5} \text{ m/s} \)
Now, calculate the force on each electron:
\( F_e = |q| v_d B = (1.6 \times 10^{-19} \text{ C}) \times (1.875 \times 10^{-5} \text{ m/s}) \times (0.2 \text{ T}) \)
\( F_e = 6 \times 10^{-25} \text{ N} \)
Thus, each electron experiences a very small but non-zero force. This problem demonstrates the microscopic origin of macroscopic electromagnetic phenomena.
In simple words: a) There is no turning force (torque) because the coil is already lined up perfectly with the magnetic field. b) The total push or pull on the entire coil is zero because it is in a steady magnetic field. c) Each tiny electron inside the wire still feels a small push from the magnetic field, which we calculated to be \( 6 \times 10^{-25} \text{ N} \).

🎯 Exam Tip: Be careful with the angle \( \theta \) for torque (between dipole moment and B) versus \( \phi \) for individual electron force (between drift velocity and B). Ensure unit conversions from cm\(^2\) to m\(^2\) are done correctly.

 

Question 4. A bar magnet is placed in a uniform magnetic field whose strength is 0.8 T. If the bar magnet is oriented at an angle 30° with the external field experiences a torque of 0.2 Nm. Calculate
i) the magnetic moment of the magnet
ii) the work is done by the magnetic field in moving it from the most stable configuration to the most its configuration and also compute the work done by the applied magnetic field in this case.
Answer: Given:
Magnetic field strength \( B = 0.8 \text{ T} \)
Angle \( \theta = 30^\circ \)
Torque \( \tau = 0.2 \text{ Nm} \)

i) Magnetic moment of the magnet (\( p_m \)):
The torque experienced by a magnetic dipole in a magnetic field is given by \( \tau = p_m B \sin\theta \).
We can rearrange this to find the magnetic moment \( p_m \):
\( p_m = \frac{\tau}{B \sin\theta} = \frac{0.2 \text{ Nm}}{(0.8 \text{ T}) \times \sin(30^\circ)} \)
\( p_m = \frac{0.2}{0.8 \times 0.5} = \frac{0.2}{0.4} = 0.5 \text{ Am}^2 \)
So, the magnetic moment of the magnet is \( 0.5 \text{ Am}^2 \).

ii) Work done by the magnetic field and by the applied magnetic field:
The potential energy of a magnetic dipole in a magnetic field is \( U = -p_m B \cos\theta \).
The most stable configuration is when \( \theta = 0^\circ \), so \( U_i = -p_m B \cos(0^\circ) = -p_m B \).
The final configuration is at \( \theta = 30^\circ \), so \( U_f = -p_m B \cos(30^\circ) \).
The work done BY the magnetic field \( W_{\text{field}} \) is \( W_{\text{field}} = -(U_f - U_i) = U_i - U_f \).
\( W_{\text{field}} = -p_m B - (-p_m B \cos(30^\circ)) = -p_m B (1 - \cos(30^\circ)) \)
\( W_{\text{field}} = -(0.5 \text{ Am}^2) \times (0.8 \text{ T}) \times (1 - \frac{\sqrt{3}}{2}) \)
\( W_{\text{field}} = -0.4 \times (1 - 0.866) = -0.4 \times 0.134 = -0.0536 \text{ J} \)
The work done BY the applied magnetic field \( W_{\text{applied}} \) is equal to the change in potential energy:
\( W_{\text{applied}} = U_f - U_i \)
\( W_{\text{applied}} = -p_m B \cos(30^\circ) - (-p_m B) = p_m B (1 - \cos(30^\circ)) \)
\( W_{\text{applied}} = (0.5 \text{ Am}^2) \times (0.8 \text{ T}) \times (1 - \frac{\sqrt{3}}{2}) \)
\( W_{\text{applied}} = 0.4 \times (1 - 0.866) = 0.4 \times 0.134 = 0.0536 \text{ J} \)
So, the work done by the magnetic field is \( -0.0536 \text{ J} \), and the work done by the applied field is \( 0.0536 \text{ J} \). This calculation helps us understand the energy changes when a magnet adjusts its orientation in a field.
In simple words: i) The strength of the magnet (its magnetic moment) is calculated to be 0.5 Am\(^2\). ii) When the magnet moves from its most comfortable position to a new angle, the magnetic field itself does negative work (meaning it takes energy). However, an outside force (applied field) has to do positive work to make this happen.

🎯 Exam Tip: Clearly distinguish between work done BY the field and work done BY the applied field, as they have opposite signs. Remember that \( \sin(30^\circ) = 0.5 \) and \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \).

 

Question 5. A non-conducting sphere has a mass of 100g and a radius of 20cm. A flat compact coil of wire with turns 5 is wrapped tightly around it with each turn concentric with the sphere. This sphere is placed on an inclined plane such that the plane of the coil is parallel to the inclined plane. A uniform magnetic field of 0.5 T exists in the region in a vertically upward direction. Compute the current I required to rest the sphere in equilibrium.
Answer: Given:
Mass of sphere \( m = 100 \text{ g} = 0.1 \text{ kg} \)
Radius of sphere \( R = 20 \text{ cm} = 0.2 \text{ m} \)
Number of turns in coil \( N = 5 \)
Magnetic field \( B = 0.5 \text{ T} \) (vertically upward)
Acceleration due to gravity \( g = 10 \text{ m/s}^2 \)
Let the inclined plane make an angle \( \theta \) with the horizontal.
For the sphere to be in equilibrium, it must be in both translational and rotational equilibrium.

Translational equilibrium:
The gravitational force acting down the incline is \( mg \sin\theta \).
This force must be balanced by an upward force, which in this case is a component of the magnetic force or friction. However, the solution implies the magnetic force is what balances a component of gravity, so let's simplify based on the given solution's approach.
For simple equilibrium on an incline, the component of gravity down the slope is \( mg \sin\theta \).

Rotational equilibrium:
The torque due to gravity about the center of the sphere is \( (mg \sin\theta)R \).
The magnetic torque on the coil is \( \tau = NIAB \sin\phi \), where \( A = \pi R^2 \) is the area of the coil. The plane of the coil is parallel to the inclined plane. The magnetic field is vertically upward. If the inclined plane makes an angle \( \theta \) with the horizontal, then the normal to the coil (and thus the magnetic moment vector) also makes an angle \( \theta \) with the vertical. So, the angle \( \phi \) between the magnetic moment \( \vec{p}_m \) and the magnetic field \( \vec{B} \) is \( \theta \).
Therefore, the magnetic torque is \( \tau = NI(\pi R^2)B \sin\theta \).
For rotational equilibrium, the magnetic torque must balance the gravitational torque:
\( NI(\pi R^2)B \sin\theta = mgR \sin\theta \)
We can cancel \( R \sin\theta \) from both sides (assuming \( R \ne 0 \) and \( \sin\theta \ne 0 \), meaning there is an incline).
\( NI\pi R B = mg \)
Now, we can solve for the current \( I \):
\( I = \frac{mg}{N\pi RB} \)
Substitute the given values:
\( I = \frac{(0.1 \text{ kg}) \times (10 \text{ m/s}^2)}{(5) \times (\pi) \times (0.2 \text{ m}) \times (0.5 \text{ T})} \)
\( I = \frac{1}{0.5 \pi} = \frac{2}{\pi} \text{ A} \)
\( I \approx 0.637 \text{ A} \)
So, a current of approximately \( 0.637 \text{ A} \) is required to keep the sphere in equilibrium. This scenario illustrates how magnetic forces can be used to counteract gravity and maintain stability in inclined systems.
In simple words: To make the sphere stay still on the slope, the turning force from the electric current in the coil must perfectly balance the turning force from gravity trying to roll the sphere down. We calculated that a current of about 0.637 Amperes is needed for this balance.

🎯 Exam Tip: This problem often confuses students with the various angles. Remember that for torque calculation, the angle is between the magnetic dipole moment (normal to the coil) and the magnetic field. For equilibrium on an inclined plane, gravity's component is \( mg \sin\theta \).

 

Question 6. Calculate the magnetic field at the center of a square loop which carries a current of 1.5 A, length of each side being is 50cm.
Answer: Let the square loop have side length \( L \). The current flowing through the loop is \( I \).
Given:
Current \( I = 1.5 \text{ A} \)
Side length \( L = 50 \text{ cm} = 0.5 \text{ m} \)
The magnetic field at the center of a square loop is the sum of the magnetic fields produced by each of its four sides. For a straight wire of finite length, the magnetic field at a point at a perpendicular distance \( r \) from its midpoint is given by \( B = \frac{\mu_0 I}{4\pi r} (\sin\theta_1 + \sin\theta_2) \).
For a square loop, consider one side. The distance from the center of the square to the midpoint of one side is \( r = \frac{L}{2} \).
From the center, the angles subtended by the ends of one side are \( \theta_1 = 45^\circ \) and \( \theta_2 = 45^\circ \) because it's a square, and the lines to the corners from the center form \( 45^\circ \) angles with the perpendicular bisector.
So, for one side:
\( B_{\text{side}} = \frac{\mu_0 I}{4\pi (L/2)} (\sin 45^\circ + \sin 45^\circ) \)
\( B_{\text{side}} = \frac{\mu_0 I}{2\pi L} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{2\pi L} (\frac{2}{\sqrt{2}}) = \frac{\mu_0 I}{\sqrt{2}\pi L} \)
Since there are four identical sides, and the magnetic field from each side points in the same direction at the center (into or out of the page, depending on the current direction), the total magnetic field at the center is four times \( B_{\text{side}} \):
\( B_{\text{total}} = 4 \times B_{\text{side}} = 4 \times \frac{\mu_0 I}{\sqrt{2}\pi L} = \frac{4\mu_0 I}{\sqrt{2}\pi L} \)
Now, substitute the values: \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \).
\( B_{\text{total}} = \frac{4 \times (4\pi \times 10^{-7} \text{ Tm/A}) \times (1.5 \text{ A})}{\sqrt{2} \times \pi \times (0.5 \text{ m})} \)
\( B_{\text{total}} = \frac{4 \times 4 \times 1.5 \times 10^{-7}}{\sqrt{2} \times 0.5} = \frac{24 \times 10^{-7}}{0.707} \)
\( B_{\text{total}} \approx 33.94 \times 10^{-7} \text{ T} \approx 3.394 \times 10^{-6} \text{ T} \approx 3.4 \times 10^{-6} \text{ T} \)
The magnetic field at the center of the square loop is approximately \( 3.4 \times 10^{-6} \text{ T} \). This calculation highlights how geometric shapes influence the resulting magnetic fields.
In simple words: To find the magnetic field at the very middle of a square wire loop, we add up the fields from each of its four sides. Using the given current and side length, we calculated the total magnetic field to be about 3.4 microteslas.

🎯 Exam Tip: Remember to calculate the field for one side first and then multiply by four. Pay attention to the angle subtended by the ends of the wire segment at the point of interest. Here, for a square's center, it's always \( 45^\circ \).

Part II:

I. Choose the incorrect pair:

 

Question 1.
1. Aurora borealis – northern lights
2. Aurora australis – southern lights
3. Bluish-Orange – collision
4. Purplish – red – Nitrogen molecules
Answer: Bluish orange – collision
In simple words: The pair "Bluish-Orange – collision" is incorrect. The bluish-orange color in auroras is usually from nitrogen molecules, not just from a general collision.

🎯 Exam Tip: For "incorrect pair" questions, identify the property that does not match the description. In aurora, colors like greenish-yellow are from oxygen, while purplish-red and bluish-orange are from nitrogen.

 

Question 2.
1. Anticlockwise current – North pole
2. Clockwise current – South pole
3. Angular momentum – \( \frac{\mathrm{nh}}{2 \pi} \)
4. Bohr magneton – \( \frac{4 \pi \mathrm{m}}{\mathrm{eh}} \)
Answer: Bohr magneton – \( \frac{4 \pi \mathrm{m}}{\mathrm{eh}} \)
In simple words: The last pair is incorrect. The Bohr magneton, which is a unit for magnetic moment, is correctly represented as \( \frac{eh}{4\pi m} \), not \( \frac{4\pi m}{eh} \).

🎯 Exam Tip: Recall the standard formula for Bohr magneton; it involves Planck's constant (h), electron charge (e), and electron mass (m), with e and h in the numerator and 4πm in the denominator.

 

Question 3.
1. Tangent law – B tan \( \theta \)
2. Tangent galvanometer – Joule’s law
3. Magnetic polarity – End rule
4. Ampere’s – circuital law – Biot – Savart law
Answer: Tangent galvanometer – Joule’s law
In simple words: The second pair is incorrect. A tangent galvanometer is used to measure current, and its working principle is based on the tangent law of magnetism, not Joule's law of heating effects.

🎯 Exam Tip: Know the fundamental laws and principles associated with common physics instruments and concepts. Joule's law relates to heat produced by current, not magnetic deflection.

II. Choose the odd man out:

 

Question 1. High permeability, high retentivity, low coercivity, thin hysteresis loop.
Answer: High retentivity – property of permanent magnet. The others (high permeability, low coercivity, thin hysteresis loop) are typically properties of electromagnets or soft magnetic materials. Having high retentivity is about holding magnetism well, which is not usually grouped with the other three characteristics for soft materials.
In simple words: "High retentivity" is the odd one out because it describes how well a magnet stays magnetized, which is good for permanent magnets. The other terms describe materials that are easy to magnetize and demagnetize, like those used in temporary magnets or electromagnets.

🎯 Exam Tip: Understand the definitions of retentivity (ability to retain magnetism), coercivity (resistance to demagnetization), permeability (how easily a material is magnetized), and hysteresis loop properties. These terms differentiate soft and hard magnetic materials.

 

Question 2. William Gilbert, Gover, Hans Christian Oersted, George Simon Ohm.
Answer: George Simon Ohm – Electricity. The others (William Gilbert, Gover, Hans Christian Oersted) are primarily known for their contributions and explanations related to magnetism or Earth's magnetic field. Ohm is famous for Ohm's law, which links voltage, current, and resistance in electrical circuits.
In simple words: George Simon Ohm is different from the others because he is famous for his work on electricity (Ohm's Law), while the other three scientists made important discoveries about magnetism.

🎯 Exam Tip: Be familiar with key scientists and their primary fields of contribution in physics, especially when distinguishing between electricity and magnetism.

 

Question 3. Aluminium, Platinum, Chromium, Bismuth
Answer: Bismuth – Diamagnetic. The others (Aluminium, Platinum, Chromium) are paramagnetic materials. Diamagnetic materials are weakly repelled by a magnetic field, whereas paramagnetic materials are weakly attracted.
In simple words: Bismuth is the odd one out because it is repelled by magnets (diamagnetic), while aluminum, platinum, and chromium are all slightly attracted to magnets (paramagnetic).

🎯 Exam Tip: Know the classification of materials based on their magnetic properties: diamagnetic (repelled), paramagnetic (weakly attracted), and ferromagnetic (strongly attracted).

 

Question 4. Phosphor bronze, suspension strip, torsion head, Dees
Answer: Dees – Parts of the cyclotron. The others (Phosphor bronze, suspension strip, torsion head) are common components found in a moving coil galvanometer. Cyclotrons are particle accelerators, which are very different from galvanometers that measure current.
In simple words: "Dees" is the odd one out because it is a part of a cyclotron, which is a machine to speed up particles. The other parts like "phosphor bronze," "suspension strip," and "torsion head" are all used in a galvanometer, which measures electric current.

🎯 Exam Tip: Understand the main components and working principles of key instruments like galvanometers and cyclotrons to differentiate their parts.

III. Fill in the blanks:

 

Question 1. When a charged particle enters in a uniform magnetic field, its kinetic energy ________________
Answer: remain constant.
In simple words: A charged particle entering a steady magnetic field will keep its speed and energy the same.

🎯 Exam Tip: Remember that the magnetic Lorentz force is always perpendicular to the velocity, so it does no work, and thus kinetic energy remains constant.

I. Multiple choice questions:

 

Question 1. The magnetic field at the center O of the following current loop is

(a) \( \frac{\mu_{0} I}{4 r} \otimes \)
(b) \( \frac{\mu_{\mathrm{o}} \mathrm{I}}{4 \mathrm{r}} \odot \)
(c) \( \frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{r}} \otimes \)
(d) \( \frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{r}} \odot \)
Answer: (a) \( \frac{\mu_{0} I}{4 r} \otimes \)
In simple words: The magnetic field at the center of a current loop is calculated using a specific formula. The direction of the field, indicated by the circle with a cross (⊗), means it points into the page.

🎯 Exam Tip: Remember the right-hand thumb rule for determining the direction of the magnetic field in a current loop; a cross indicates into the page, and a dot indicates out of the page.

 

Question 2. An electron moves straight line inside a charged parallel plate capacitor of uniform charge density \( \sigma \). The time taken by the electron to cross the parallel plate capacitor undeflected when the plates of the capacitor are kept under constant magnetic field of induction B is

(a) \( \varepsilon_{0} \frac{\mathrm{elB}}{\sigma} \)
(b) \( \varepsilon_{\mathrm{O}} \frac{\mathrm{lB}}{\sigma l} \)
(c) \( \varepsilon_{\mathrm{o}} \frac{\mathrm{lB}}{\mathrm{e} \sigma} \)
(d) \( \varepsilon_{0} \frac{\mathrm{B}}{\sigma} \)
Answer: (a) \( \varepsilon_{0} \frac{\mathrm{elB}}{\sigma} \)
In simple words: For an electron to pass through a capacitor and a magnetic field without changing direction, the electric force must balance the magnetic force. The time it takes depends on the distance, the electric field, and the magnetic field.

🎯 Exam Tip: For undeflected motion in crossed electric and magnetic fields, the electric force (qE) must be equal and opposite to the magnetic force (qvB). This balance is key to solving such problems.

 

Question 3. A particle having mass m and charge q accelerated through a potential difference V. Find the force experienced when it is kept under perpendicular magnetic field B is

(a) \( \sqrt{\frac{2 q^{3} B V}{m}} \)
(b) \( \sqrt{\frac{q^{3} B^{2} V}{2 m}} \)
(c) \( \sqrt{\frac{2 q^{3} B^{2} V}{m}} \)
(d) \( \sqrt{\frac{2 q^{3} B V}{m^{3}}} \)
Answer: (c) \( \sqrt{\frac{2 q^{3} B^{2} V}{m}} \)
In simple words: When a charged particle is accelerated by voltage and then moves in a magnetic field, it experiences a force. This force depends on its charge, the magnetic field strength, and how fast it is moving after being accelerated.

🎯 Exam Tip: To solve this, first find the velocity (v) of the charged particle using the kinetic energy gained from the potential difference (\( \frac{1}{2} mv^2 = qV \)). Then, substitute this velocity into the formula for magnetic force (F = qvB).

 

Question 4. A circular coil of radius 5 cm and has 50 turns carries a current of 3 amperes. The magnetic dipole moment of the coil is
(a) 1.0 Am\(^2\)
(b) 1.2 Am\(^2\)
(c) 0.5 Am\(^2\)
(d) 0.8 Am\(^2\)
Answer: (b) 1.2 Am\(^2\)
In simple words: The magnetic dipole moment tells us how strong a magnet a current loop acts like. It depends on the number of turns, the current flowing, and the area of the coil.

🎯 Exam Tip: Always convert radius from cm to meters (m) before using it in calculations. The magnetic dipole moment (M) for a coil is given by M = NIA, where N is the number of turns, I is the current, and A is the area of the coil.

 

Question 5. A thin insulated wire forms a plane spiral of N=100 tight turns carrying a current I = 8mA (milliampere). The radii of inside and outside turns are a = 50mm and b = 100 mm, respectively. The magnetic induction at the center of the spiral is
(a) 5 µT
(b) 7 µT
(c) 8 µT
(d) 10 µT
Answer: (b) 7 µT
In simple words: A spiral with current makes a magnetic field in the middle. The strength of this field depends on how many turns there are, how much current flows, and the sizes of the inner and outer parts of the spiral.

🎯 Exam Tip: When dealing with spirals, remember that the magnetic field is not uniform across the entire spiral. For such problems, integrating the field contributions from infinitesimal turns is often required, or using a specific formula for a flat spiral if available.

 

Question 6. Three wires of equal lengths are bent in the form of loops. One of the loops is a circle, another is a semi-circle and the third one is a square. They are placed in a uniform magnetic field and the same electric current is passed through them. Which of the following loop configuration will experience greater torque?
(a) circle
(b) semi-circle
(c) square
(d) all of them
Answer: (a) Circle
In simple words: When different shapes of wires carrying current are placed in a magnetic field, the circle shape feels the strongest twisting force. This is because, for the same length of wire, a circle encloses the biggest area.

🎯 Exam Tip: The torque on a current loop in a magnetic field is directly proportional to the area enclosed by the loop. For a given perimeter (length of wire), a circle encloses the maximum possible area.

 

Question 7. Two identical coils, each with N turns and radius R are placed coaxially at a distance R as shown in the figure. If I is the current passing through the loops in the same direction, then the magnetic field at a point P at a distance of R/2 from the center of each coil

(a) \( \frac{8 \mathrm{~N} \mu_{\mathrm{o}} \mathrm{I}}{\sqrt{5} \mathrm{R}} \)
(b) \( \frac{8 \mathrm{~N} \mu_{\mathrm{o}} \mathrm{I}}{5^{3 / 2} \mathrm{R}} \)
(c) \( \frac{8 \mathrm{~N} \mu_{\mathrm{O}} \mathrm{I}}{5 \mathrm{R}} \)
(d) \( \frac{4 \mathrm{~N} \mu_{\mathrm{o}} \mathrm{I}}{\sqrt{5} \mathrm{R}} \)
Answer: (b) \( \frac{8 \mathrm{~N} \mu_{\mathrm{o}} \mathrm{I}}{5^{3 / 2} \mathrm{R}} \)
In simple words: When two coils are placed close together with current flowing, they create a magnetic field. We are looking for the total magnetic field at a specific point exactly in the middle of these two coils.

🎯 Exam Tip: This setup is a Helmholtz coil arrangement. The total magnetic field at the midpoint is the sum of the fields produced by each coil individually. Remember the formula for the magnetic field on the axis of a circular coil.

 

Question 8. A wire of length l carrying a current I along the Y direction is kept in a and magnetic field is given by \( \mathrm{B}=\frac{\beta}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k}) \) T. The magnitude of Lorentz force acting on the wire is
(a) \( \sqrt{\frac{2}{\sqrt{3}}} \beta \mathrm{I} l \)
(b) \( \sqrt{\frac{1}{\sqrt{3}}} \beta I l \)
(c) \( \sqrt{2}\beta Il \)
(d) \( \sqrt{\frac{1}{2}} \beta \mathrm{I} l \)
Answer: (a) \( \sqrt{\frac{2}{\sqrt{3}}} \beta \mathrm{I} l \)
In simple words: When a current-carrying wire is placed in a magnetic field, it experiences a force. The size of this force depends on the current, the length of the wire, and the strength and direction of the magnetic field.

🎯 Exam Tip: The force on a current-carrying wire in a magnetic field is given by \( \vec{F} = I (\vec{l} \times \vec{B}) \). Pay close attention to the vector cross product, especially when the current direction and magnetic field have multiple components.

 

Question 9. A bar magnet of length l and magnetic moment \( p_m \) is bent in the form of an arc as shown in Figure. The new magnetic dipole moment will be

(a) \( p_m \)
(b) \( \frac{3}{\pi} p_{m} \)
(c) \( \frac{2}{\pi} \mathrm{p}_{\mathrm{m}} \)
(d) \( \frac{1}{2} \mathrm{p}_{\mathrm{m}} \)
Answer: (b) \( \frac{3}{\pi} p_{m} \)
In simple words: When you bend a straight magnet into a curve, its magnetic moment changes. The magnetic moment is like a measure of its overall magnetic strength and depends on the effective distance between its poles.

🎯 Exam Tip: The magnetic moment of a bar magnet is \( p_m = q_m \times l \), where \( q_m \) is the pole strength and \( l \) is the magnetic length. When bent into an arc, the magnetic length becomes the straight-line distance between the new pole positions, which is less than the original length. For a bent magnet, the effective length is \( 2r \sin(\theta/2) \) where \( \theta \) is the angle subtended by the arc at the center.

 

Question 10. A non-conducting charged ring of charge of q, mass m and radius r is rotated about its axis with constant angular speed to. Find the ratio of its magnetic moment with angular momentum is
(a) \( \frac{q}{m} \)
(b) \( \frac{2 q}{m} \)
(c) \( \frac{\mathrm{q}}{2 \mathrm{~m}} \)
(d) \( \frac{q}{4 m} \)
Answer: (c) \( \frac{\mathrm{q}}{2 \mathrm{~m}} \)
In simple words: When a charged ring spins, it creates both a magnetic moment and angular momentum. The relationship between these two is always a fixed value, known as the gyromagnetic ratio.

🎯 Exam Tip: For a charged rotating body, the gyromagnetic ratio (ratio of magnetic moment to angular momentum) is always constant and equal to \( \frac{q}{2m} \), where q is the charge and m is the mass of the body.

 

Question 11. The BH curve for a ferromagnetic material is shown in the Figure. The material is placed inside a long solenoid which contains 1000 turns/cm. The current that should be passed in the solenoid to demagnetize the ferromagnet completely is

(a) 1.00 mA
(b) 1.25 mA
(c) 1.50 mA
(d) 1.75 mA
Answer: (c) 1.50 mA
In simple words: To remove the magnetism from a ferromagnetic material, you need to apply a magnetic field in the opposite direction. The graph shows how much reverse magnetic field (H) is needed to bring the magnetic induction (B) back to zero.

🎯 Exam Tip: The coercivity (the magnetic field H needed to demagnetize the material, i.e., to bring B to zero) is found at the point where the hysteresis loop crosses the H-axis. Use the relation H = nI to find the current I, where n is turns per unit length.

 

Question 12. Two short bar magnets have magnetic moments 1.20 Am\(^2\) and 1.00 Am\(^2\), respectively. They are kept on a horizontal table parallel to each other with their north poles pointing towards the south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centers is (Horizontal components of Earth’s magnetic induction is 3.6 × 10\(^{-5}\) Wbm\(^{-2}\))
(a) 3.60 × 10\(^{-5}\) Wbm\(^{-2}\)
(b) 3.5 × 10\(^{-5}\) Wbm\(^{-2}\)
(c) 2.56 × 10\(^{-4}\) Wbm\(^{-2}\)
(d) 2.2 × 10\(^{-4}\) Wbm\(^{-2}\)
Answer: (c) 2.56 × 10\(^{-4}\) Wbm\(^{-2}\)
In simple words: Two small magnets are placed side-by-side, pointing in the same direction. We need to find the total magnetic field strength right in the middle of them, also considering the Earth's magnetic field.

🎯 Exam Tip: The magnetic field on the equatorial line of a bar magnet is given by \( B = \frac{\mu_0}{4\pi} \frac{M}{(d^2+L^2)^{3/2}} \). When calculating the resultant field, add the contributions from both magnets (considering their directions) and then add the Earth's horizontal component, also accounting for direction.

 

Question 13. The vertical component of Earth’s magnetic field at a place is equal to the horizontal component. What is the value of the angle of dip at this place?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer: (b) 45°
In simple words: The Earth's magnetic field has both an up-and-down part (vertical) and a side-to-side part (horizontal). If these two parts are equally strong, the magnetic needle will dip at a 45-degree angle.

🎯 Exam Tip: The angle of dip (I) is related to the vertical (Bv) and horizontal (BH) components of Earth's magnetic field by \( \tan I = \frac{B_V}{B_H} \). If \( B_V = B_H \), then \( \tan I = 1 \), which means \( I = 45^\circ \).

 

Question 14. A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is \( \sigma \). The disc rotates about an axis perpendicular to its plane passing through the center with angular velocity \( \omega \). Find the magnitude of the torque on the disc if it is placed in a uniform magnetic field whose strength is B which is directed perpendicular to the axis of rotation
(a) \( \frac{1}{4}\sigma \pi BR \)
(b) \( \frac{1}{4}\sigma \pi BR^2 \)
(c) \( \frac{1}{4}\sigma \pi BR^3 \)
(d) \( \frac{1}{4}\sigma \pi BR^4 \)
Answer: (d) \( \frac{1}{4}\sigma \pi BR^4 \)
In simple words: When a charged disc spins in a magnetic field, it experiences a twisting force (torque). This force depends on how much charge is on the disc, how fast it spins, its size, and the strength of the magnetic field.

🎯 Exam Tip: The torque on a magnetic dipole in an external magnetic field is given by \( \tau = MB \sin\theta \). First, calculate the magnetic moment (M) of the rotating charged disc, which is \( M = \frac{1}{4} q \omega R^2 \). Remember to integrate for the entire disc if charge density is uniform.

 

Question 15. The potential energy of magnetic dipole whose dipole moment is \( \vec{p}_{m}= (-0.5 \hat{i}+0.4 \hat{j}) \mathrm{Am}^{2} \) Am kept in uniform magnetic field \( \vec{B}=0.2 \hat{i} \mathrm{~T} \)
(a) -0.1 J
(b) -0.8 J
(c) 0.1 J
(d) 0.8 J
Answer: (a) -0.1 J
In simple words: A magnetic dipole, like a small magnet, has stored energy when placed in a magnetic field. This energy depends on how the dipole is aligned with the field and can be calculated using their vector dot product.

🎯 Exam Tip: The potential energy (U) of a magnetic dipole in a magnetic field is given by \( U = -\vec{p}_m \cdot \vec{B} \). Perform the dot product of the given magnetic moment and magnetic field vectors carefully.

 

Question 17. Dimension of Resistance is ______.
(a) \( \text{ML}^2 \text{T}^{-3} \text{A}^{-2} \)
(b) \( \text{ML}^2 \text{T}^{-1} \text{A}^{-1} \)
(c) \( \text{ML}^2 \text{T}^2 \text{A}^{-3} \)
(d) \( \text{ML}^2 \text{T}^{-1} \text{A}^{-2} \)
Answer: (a) \( \text{ML}^2 \text{T}^{-3} \text{A}^{-2} \)
In simple words: The correct dimension for resistance is found by using Ohm's Law, V=IR, and then expressing voltage and current in terms of fundamental units like mass (M), length (L), time (T), and current (A).

🎯 Exam Tip: Remember fundamental physical quantities and their dimensions to easily derive the dimensions of other physical quantities like resistance.

 

Question 18. Photo has spin _______.
(a) \( S = \frac{1}{2} \)
(b) \( S = -\frac{1}{2} \)
(c) S = 1
(d) S = 2
Answer: (c) S = 1
In simple words: A photon, which is a particle of light, has a spin quantum number of 1. This means it behaves in a certain way that affects its properties like polarization.

🎯 Exam Tip: Understanding the spin of fundamental particles like photons is key in quantum mechanics and optics, influencing how they interact with matter.

 

Question 19. At Curie point, a ferromagnetic material becomes-
(a) non magnetic
(b) diamagnetic
(c) paramagnetic
(d) antiferromagnetic
Answer: (c) paramagnetic
In simple words: When a ferromagnetic material is heated past its Curie point, it loses its strong magnetic properties and becomes paramagnetic. This is because the thermal energy disrupts the alignment of magnetic domains.

🎯 Exam Tip: The Curie point is a critical temperature for ferromagnetic materials; exceeding it causes them to lose their permanent magnetism and transition to a weaker paramagnetic state.

 

Question 20. The variation of magnetic susceptibility (k) with magnetizing field (H) for a paramagnetic substance is _______

Answer:
The magnetic susceptibility of a paramagnetic substance decreases as the magnetizing field (H) increases, but the relationship is not perfectly linear. The graph shows a decreasing curve, indicating that the material becomes less susceptible to magnetization as the field gets stronger, or rather, the effect of the field diminishes relative to the material's properties at higher fields.
In simple words: For a paramagnetic material, as the magnetic field gets stronger, its magnetic susceptibility goes down. This means it becomes less easy to magnetize it further with a stronger field.

🎯 Exam Tip: Remember that paramagnetic materials have positive but small susceptibility which depends on temperature and field strength, distinguishing them from diamagnetic and ferromagnetic substances.

 

Question 21. The force per unit length between two conductors carrying one ampere current each and separated by a distance of 1 m is
(a) \( 2\pi \times 10^{-7} \text{ N} \)
(b) \( 2 \times 10^{-7} \text{ N} \)
(c) \( 4\pi \times 10^7 \text{ N} \)
(d) \( 2\pi \times 10^7 \text{ N} \)
Answer: (b) \( 2 \times 10^{-7} \text{ N} \)
In simple words: When two long, straight wires each carry 1 Ampere of current and are 1 meter apart, the magnetic force between them per unit length is defined as \( 2 \times 10^{-7} \text{ Newtons} \). This value is used to define the Ampere unit itself.

🎯 Exam Tip: This specific force value is the definition of the Ampere in the SI system, making it a crucial fundamental constant in electromagnetism.

 

Question 22. A cyclotron cannot accelerate _______.
(a) electrons
(b) protons
(c) deuterons
(d) \( \alpha \) – particles
Answer: (a) electrons
In simple words: Cyclotrons are designed to speed up heavy charged particles like protons, not very light particles like electrons. Electrons are too light, so they quickly gain speed and their mass changes significantly due to relativity, which makes the cyclotron's timing go wrong.

🎯 Exam Tip: The main limitation for accelerating electrons in a cyclotron is their small mass, which leads to relativistic effects becoming significant at much lower energies, disrupting the cyclotron's resonant frequency condition.

 

Question 23. \( \oint \overrightarrow{\mathbf{B}} \cdot \overrightarrow{\mathbf{d}}l \) is equal to
(a) \( \mu_0 \text{I} \)
(b) \( \mu_0 \text{ I}_0 \)
(c) \( \frac{\mu_0}{\text{I}_0} \)
(d) \( \frac{\text{I}_0}{\mu_0} \)
Answer: (b) \( \mu_0 \text{ I}_0 \)
In simple words: Ampere's law states that the closed loop integral of the magnetic field (B) dotted with the path element (dl) is equal to the permeability of free space (\( \mu_0 \)) multiplied by the total current (\( \text{I}_0 \)) passing through the loop.

🎯 Exam Tip: This equation is Ampere's Circuital Law, a fundamental principle linking electric currents to the magnetic fields they create.

 

Question 24. A magnetic needle is placed in a uniform magnetic field. It experience-
(a) a force and a torque
(b) a force but not a torque
(c) a torque but not a force
(d) neither a force nor a torque
Answer: (c) a torque but not a force
In simple words: When a magnetic needle is put in a uniform magnetic field, the forces on its north and south poles are equal and opposite, so there's no net force moving it. However, these forces act at different points, creating a twisting effect, which is a torque, causing the needle to align with the field.

🎯 Exam Tip: A uniform magnetic field exerts only torque on a magnetic dipole (like a needle) because the net force on the poles cancels out. A non-uniform field, however, would exert both a net force and a torque.

 

Question 25. In a galvanometer, the magnetic field used is _____
(a) radial
(b) parallel
(c) perpendicular
(d) tangential
Answer: (a) radial
In simple words: Galvanometers use a special kind of magnetic field called a radial field. This field ensures that the force on the coil is always perpendicular to its sides, which makes the deflection directly proportional to the current, providing a linear scale.

🎯 Exam Tip: The use of a radial magnetic field in a moving coil galvanometer is crucial for ensuring that the torque acting on the coil is proportional to the current, leading to a linear scale for readings.

 

Question 26. In a tangent galvanometer, a current 1A produces a deflection of 30° the current required to produce a deflection of 60° is
(a) 3 A
(b) 2 A
(c) \( \sqrt{2}\text{A} \)
(d) \( \sqrt{3}\text{A} \)
Answer: (a) 3A
In simple words: In a tangent galvanometer, the current is proportional to the tangent of the deflection angle. If 1A gives \( \tan(30^\circ) \), then to get \( \tan(60^\circ) \), you need 3A because \( \tan(60^\circ) \) is \( \sqrt{3} \) and \( \tan(30^\circ) \) is \( 1/\sqrt{3} \), so \( \sqrt{3} / (1/\sqrt{3}) = 3 \).

🎯 Exam Tip: For tangent galvanometers, remember the relationship \( \text{I} \propto \tan \theta \). This means current is directly proportional to the tangent of the deflection angle, which is essential for calculations.

 

Question 27. A magnetic needle suspended freely-
(a) orients itself in a definite direction
(b) remains in any direction
(c) become vertical with N-pole up
(d) become vertical with N-pole down
Answer: (a) orients itself in a definite direction
In simple words: A magnetic needle, when allowed to move freely, will always point itself in the Earth's magnetic north-south direction. This is why compasses work.

🎯 Exam Tip: The alignment of a freely suspended magnetic needle with the Earth's magnetic field is a fundamental concept in magnetism and the basis for compass operation.

 

Question 28. The voltmeter resistance is
(a) \( \text{R}_{\text{g}} = \text{R}_{\text{v}} + \text{R}_{\text{h}} \)
(b) \( \text{R}_{\text{v}} = \text{R}_{\text{g}} + \text{R}_{\text{h}} \)
(c) \( \text{R}_{\text{h}} = \text{R}_{\text{v}} + \text{R}_{\text{g}} \)
(d) \( \text{R}_{\text{g}} = \text{R}_{\text{h}} - \text{R}_{\text{v}} \)
Answer: (b) \( \text{R}_{\text{v}} = \text{R}_{\text{g}} + \text{R}_{\text{h}} \)
In simple words: A voltmeter is made by adding a high resistance (Rh) in series with a galvanometer (Rg). So, the total resistance of the voltmeter (Rv) is the sum of these two resistances.

🎯 Exam Tip: Always remember that a voltmeter is connected in parallel to measure potential difference and thus requires very high resistance (galvanometer plus a series resistance) to draw minimal current from the circuit.

 

Question 29. Kirchoff’s I an II laws are based on conservation of
(a) charge and energy
(b) energy and charge
(c) energy and voltage
(d) energy and current
Answer: (a) charge and energy
In simple words: Kirchhoff's Current Law (KCL) is based on the conservation of electric charge, meaning charge cannot be created or destroyed at a junction. Kirchhoff's Voltage Law (KVL) is based on the conservation of energy, meaning energy is conserved around any closed loop in a circuit.

🎯 Exam Tip: KCL (current) is about charge conservation at a junction, and KVL (voltage) is about energy conservation around a loop. Confusing these is a common mistake.

 

Question 30. A moving charge produces-
(a) an electric field only
(b) a magnetic field only
(c) both electric and magnetic fields
(d) neither an electric nor a magnetic field
Answer: (c) both electric and magnetic fields
In simple words: A charge that is moving creates both an electric field around it (because it is a charge) and a magnetic field (because it is a moving charge, which is essentially a current).

🎯 Exam Tip: Remember that stationary charges produce only electric fields, but moving charges produce both electric and magnetic fields. This fundamental concept is central to electromagnetism.

XI. Two Mark Questions:

 

Question 1. Define magnetic declination (D).
Answer: Magnetic declination (D) is the angle between the true geographical meridian and the magnetic meridian at a specific location on Earth. It tells you how far off your compass reading is from true north.
In simple words: Magnetic declination is the angle difference between where your compass points (magnetic north) and the actual North Pole (true north).

🎯 Exam Tip: State that magnetic declination is the angle between the geographical and magnetic meridians. This angle varies with location and is important for accurate navigation.

 

Question 2. What is a magnetizing field?
Answer: A magnetizing field is an external magnetic field applied to a material to magnetize it. This field is a vector quantity and is denoted by \( \overrightarrow{\text{H}} \). Its unit is Ampere per meter (Am\(^{-1}\)).
In simple words: A magnetizing field is the outside magnetic force used to make something magnetic. It is measured in Ampere per meter.

🎯 Exam Tip: Define the magnetizing field as the external magnetic field that induces magnetization in a material and state its vector nature and unit.

 

Question 3. What is relative permeability?
Answer: Relative permeability (\( \mu_{\text{r}} \)) is the ratio of the absolute permeability of a material (\( \mu \)) to the permeability of free space (\( \mu_0 \)). For free space, \( \mu_{\text{r}} = 1 \). If a material is isotropic, \( \mu \) is a scalar quantity; otherwise, it can be a tensor. This value shows how much better a material is at letting magnetic field lines pass through it compared to a vacuum.
In simple words: Relative permeability shows how easily magnetic lines can pass through a material compared to how easily they pass through empty space. It is a ratio of the material's magnetic permeability to that of a vacuum.

🎯 Exam Tip: Define relative permeability as a ratio and mention its value for free space. This concept is crucial for understanding how different materials interact with magnetic fields.

 

Question 4. What is the intensity of magnetisation?
Answer: The intensity of magnetization (\( \overrightarrow{\text{M}} \)) is defined as the net magnetic moment per unit volume of a material. It represents how strongly a material is magnetized. Its unit is Ampere per meter (Am\(^{-1}\)).
In simple words: The intensity of magnetization tells us how much magnetic strength is packed into each part of a material. It is the magnetic moment per unit volume.

🎯 Exam Tip: Clearly state that magnetization intensity is the net magnetic moment per unit volume. This quantifies the magnetic state of a material.

 

Question 5. State Curie’s law.
Answer: Curie’s law states that for a paramagnetic material, the magnetic susceptibility (\( \text{X}_{\text{m}} \)) is inversely proportional to its absolute temperature (T). This can be expressed as \( \text{X}_{\text{m}} = \frac{\text{C}}{\text{T}} \), where C is the Curie constant. This means as the temperature of the material increases, its susceptibility to magnetism decreases.
In simple words: Curie's law says that a material's magnetic susceptibility goes down as its temperature goes up. The hotter it gets, the less easily it can be magnetized.

🎯 Exam Tip: Mention the inverse relationship between magnetic susceptibility and absolute temperature, and include the formula with the Curie constant.

 

Question 6. Define in terms of magnetising field.
Answer: The magnetizing field (\( \overrightarrow{\text{H}} \)) is the external magnetic field used to magnetize a sample or specimen. It is a vector quantity, meaning it has both magnitude and direction, and its unit is Ampere per meter (A \( \text{m}^{-1} \)). It is the field that causes the magnetic dipoles within a material to align.
In simple words: The magnetizing field is the outside magnetic force that makes a material magnetic. It is a vector and is measured in Ampere per meter.

🎯 Exam Tip: Define the magnetizing field as the external field causing magnetization, highlight its vector nature, and state its unit clearly.

 

Question 7. What kind of magnetic properties electromagnet has?
Answer: Materials suitable for electromagnets should possess the following magnetic properties:

• High initial permeability
• Thin hysteresis loop
• Low retentivity
• Low coercivity

Examples include soft iron and Mumetal (a nickel-iron alloy). These properties allow the electromagnet to be easily magnetized and demagnetized.
In simple words: Electromagnets need materials that can be easily made magnetic and then unmade magnetic. This means they need high initial permeability, a thin hysteresis loop, and low retentivity and coercivity.

🎯 Exam Tip: List at least three key properties (e.g., high permeability, low retentivity, low coercivity) and give an example of a suitable material like soft iron.

 

Question 8. What is needed to design transformer cores?
Answer: To design transformer cores, materials with the following magnetic properties are needed:
1. Large magnetic induction
2. High initial permeability
3. Thin hysteresis loop with a small area
These properties ensure that the core can easily guide magnetic flux and minimize energy loss during magnetization and demagnetization cycles, making the transformer efficient.
In simple words: Transformer cores need materials that can create strong magnetic fields, let magnetic lines pass through them easily, and waste very little energy when they are magnetized and demagnetized.

🎯 Exam Tip: When discussing transformer cores, emphasize properties that lead to high efficiency and minimal energy loss, such as high permeability and a small hysteresis loop area.

 

Question 9. State ‘One ampere’.
Answer: One ampere is defined as the constant current that, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed one meter apart in a vacuum, would produce between these conductors a force equal to \( 2 \times 10^{-7} \) newton per meter of length. This definition is fundamental to the SI unit system.
In simple words: One Ampere is a specific amount of electricity flow. It is defined by the magnetic force created between two long wires carrying this current, placed 1 meter apart in empty space.

🎯 Exam Tip: Accurately state the force value and the conditions (parallel conductors, 1m apart, vacuum, infinite length) for defining one Ampere.

 

Question 10. Define magnetic dipole moment of the current loop.
Answer: The magnetic dipole moment of a current loop is defined as the product of the current flowing through the loop (I) and the area of the loop (A). It is a vector quantity, with its direction given by the right-hand thumb rule. This moment measures the strength and orientation of a magnetic source.
In simple words: The magnetic dipole moment of a current loop is found by multiplying the current flowing in the loop by the area it covers. This tells you how strong its magnetic field is.

🎯 Exam Tip: Define the magnetic dipole moment as \( \text{M} = \text{IA} \) (or \( \overrightarrow{\text{M}} = \text{I}\overrightarrow{\text{A}} \)), mentioning both the magnitude (product of current and area) and direction (right-hand rule).

 

Question 11. What is a solenoid?
Answer: A solenoid is a long coil of insulated wire closely wound in the form of a helix. When an electric current passes through the wire, it creates a nearly uniform magnetic field inside the coil, similar to that of a bar magnet. This design makes solenoids useful in many electrical applications.
In simple words: A solenoid is like a spring-shaped coil of wire. When electricity flows through it, it acts like a magnet, making a strong magnetic field inside.

🎯 Exam Tip: Describe a solenoid as a closely wound helical coil and mention its ability to produce a strong, nearly uniform magnetic field when current flows through it.

 

Question 12. What is toroid?
Answer: A toroid is essentially a solenoid that has been bent into a closed ring shape, joining its two ends. This configuration results in a magnetic field that is confined almost entirely within the toroid itself, with virtually no magnetic field outside or inside the empty space of the ring. It is used in applications where a closed magnetic flux path is required.
In simple words: A toroid is a wire coil shaped like a donut. Its magnetic field is mostly kept inside the donut shape, with almost no field outside.

🎯 Exam Tip: Explain a toroid as a solenoid bent into a closed ring, and note that its magnetic field is primarily confined within the ring structure.

 

Question 13. Define one tesla.
Answer: One Tesla (1 T) is defined as the strength of a magnetic field such that a unit charge (1 Coulomb) moving with a unit velocity (1 meter per second) perpendicular to the magnetic field experiences a unit force (1 Newton). In simpler terms, \( 1 \text{ T} = 1 \text{ Ns/Cm} = 1 \text{ N/(Am)} = 1 \text{ NA}^{-1}\text{m}^{-1} \). This is a very strong unit of magnetic field strength.
In simple words: One Tesla is the amount of magnetic field strength that makes a 1 Coulomb charge, moving 1 meter per second, feel a 1 Newton force when moving straight across the field.

🎯 Exam Tip: Provide the definition of Tesla based on the Lorentz force on a moving charge. Also, mentioning its equivalence in other units like N/(Am) is beneficial.

 

Question 14. State Lorentz force.
Answer: The Lorentz force is the total force experienced by a charged particle moving in both electric and magnetic fields. It is given by the formula \( \overrightarrow{\text{F}} = \text{q}\overrightarrow{\text{E}} + \text{q}(\overrightarrow{\text{V}} \times \overrightarrow{\text{B}}) \). This force is fundamental to understanding how charged particles behave in electromagnetic fields.
In simple words: Lorentz force is the total force a charged particle feels when it is moving through both electric and magnetic fields.

🎯 Exam Tip: State the complete vector form of the Lorentz force, showing the contributions from both the electric field \( (\text{q}\overrightarrow{\text{E}}) \) and the magnetic field \( (\text{q}(\overrightarrow{\text{V}} \times \overrightarrow{\text{B}})) \).

 

Question 15. Give the expression for cyclotron frequency or gyro frequency.
Answer: The cyclotron frequency (or gyro frequency) is the frequency at which a charged particle moves in a circular path when subjected to a uniform magnetic field perpendicular to its velocity. The expressions are:
Frequency: \( \text{f} = \frac{\text{qB}}{2\pi\text{m}} \)
Angular frequency: \( \omega = 2\pi\text{f} = \frac{\text{qB}}{\text{m}} \)
This frequency depends only on the charge-to-mass ratio of the particle and the strength of the magnetic field.
In simple words: Cyclotron frequency is how often a charged particle goes around in a circle inside a magnetic field. It depends on the particle's charge, its mass, and the strength of the magnetic field.

🎯 Exam Tip: Provide both the linear frequency (f) and angular frequency (ω) expressions, and state that it is independent of the particle's velocity and the radius of its path.

 

Question 16. Give the uses of the mass spectrometer.
Answer: A mass spectrometer is a versatile instrument used in various fields:
1. In sciences: It is used in medicine for drug analysis, space science for analyzing atmospheric compositions, and geology for isotope dating.
2. In medicine: Anesthesiologists use it to measure respiratory gases during surgery.
3. Biologists: They use it to determine reaction mechanisms in complex processes like photosynthesis.
It helps identify unknown compounds, determine isotopic composition, and analyze molecular structures.
In simple words: A mass spectrometer is a tool used in many areas like science, medicine, and biology. It helps find out what substances are made of, like checking gasses during surgery or understanding how plants make food.

🎯 Exam Tip: Focus on the main applications like identifying unknown compounds, analyzing isotopes, and its use in medical diagnostics and scientific research.

 

Question 17. How is a galvanometer is converted into an ammeter and a voltmeter?
Answer:
1. To convert a galvanometer into an ammeter: A low resistance, called a shunt (S), is connected in parallel with the galvanometer. This allows most of the current to bypass the galvanometer, enabling it to measure larger currents.
2. To convert a galvanometer into a voltmeter: A high resistance (\( \text{R}_{\text{h}} \)) is connected in series with the galvanometer. This increases the total resistance and allows the instrument to measure large potential differences without drawing too much current from the circuit.
In simple words: To make an ammeter, you add a small resistor next to the galvanometer. To make a voltmeter, you add a large resistor in a line with the galvanometer.

🎯 Exam Tip: Clearly distinguish between the two conversions: ammeter requires low resistance in parallel (shunt), while voltmeter requires high resistance in series.

 

Question 18. What is the reason for the hemispherical magnetic poles used in moving coil galvanometer?
Answer: Hemispherical magnetic poles are used in moving coil galvanometers for two main reasons:
1. They produce a radial magnetic field: This means the magnetic field lines are always perpendicular to the plane of the coil, regardless of the coil's position within the field.
2. The plane of the coil remains parallel to the magnetic field in all its positions: This ensures that the torque on the coil is always maximum and directly proportional to the current, resulting in a linear scale for readings.
In simple words: Hemispherical magnets are used in galvanometers to create a special "radial" magnetic field. This field makes sure the coil always feels the strongest twist, so the needle moves evenly with the electric current.

🎯 Exam Tip: Emphasize that radial magnetic fields ensure that the torque is always proportional to the current, leading to a linear scale for the galvanometer.

 

Question 19. Why did we use the coils in different thicknesses in the Tangent galvanometer?
Answer: The coils in a Tangent galvanometer are designed with different thicknesses to provide varying numbers of turns (e.g., 2, 5, and 50 turns). This allows the instrument to measure currents of different strengths (ranges) accurately. By changing the coil, the galvanometer's sensitivity can be adjusted to suit the magnitude of the current being measured.
In simple words: Tangent galvanometers have coils of different thicknesses to give them different numbers of turns. This lets them measure weak currents or strong currents accurately.

🎯 Exam Tip: The use of multiple coils (different turns/thicknesses) allows the tangent galvanometer to operate over a wide range of current measurements by adjusting its sensitivity.

 

Question 20. The electrons cannot be accelerated by a cyclotron, why?
Answer: Electrons cannot be accelerated by a cyclotron for two main reasons:
1. Relativistic variation of mass: At the high frequencies and speeds achieved in a cyclotron, the mass of electrons increases significantly due to relativistic effects. This change in mass disrupts the constant frequency condition required for the cyclotron to operate efficiently.
2. High frequencies required: Due to their very small mass, electrons would require extremely high frequencies to maintain resonance, which is technically challenging to achieve.
In simple words: Cyclotrons cannot speed up electrons because electrons are so light. When they go very fast, their mass changes too much, messing up the cyclotron's timing. Also, the required speed would need extremely high frequencies that are hard to create.

🎯 Exam Tip: The key reason is the relativistic mass variation of electrons at high speeds, which violates the cyclotron's fixed-frequency operation condition.

 

Question 21. Why Phosphor bronze wire is used as suspension wire in moving coil galvanometer?
Answer: Phosphor bronze wire is used as a suspension wire in a moving coil galvanometer because:
1. It produces a small couple per unit twist: This means it requires very little force to twist it, making the galvanometer very sensitive to small currents.
2. A small couple per unit twist increases the sensitivity: The less resistance the wire offers to twisting, the more the coil will deflect for a given current, thus increasing the instrument's ability to detect tiny currents.
In simple words: Phosphor bronze wire is used because it twists very easily. This makes the galvanometer very sensitive, so it can detect even tiny electric currents.

🎯 Exam Tip: Focus on the low torsional constant (small couple per unit twist) of phosphor bronze, which directly leads to high sensitivity in the galvanometer.

 

Question 22. Why freely suspended bar magnets in your lab experience only torque but not any translatory motion even though Earth has a non-uniform magnetic field?
Answer: While the Earth's magnetic field is non-uniform on a large scale, within the small confines of a laboratory, the field can be considered approximately uniform. Therefore, a freely suspended bar magnet in the lab experiences a torque that aligns it with the local magnetic field but does not experience any net translatory force. This means it twists but does not move from its position.
In simple words: Even though Earth's magnetic field is not perfectly even everywhere, in a small lab, it is mostly uniform. So, a magnet will only twist to line up with the field, not move from its spot.

🎯 Exam Tip: The crucial point here is that Earth's magnetic field, though non-uniform globally, can be approximated as uniform over small laboratory distances, thus exerting only torque on a dipole.

 

Question 23. List the differences between coulomb force and Lorentz force.
Answer:

In simple words: Coulomb force is between two charges and gets weaker with distance, while Lorentz force is on a moving charge in a magnetic field and does not change with distance. Also, Coulomb force changes depending on the material between the charges, but Lorentz force does not.

🎯 Exam Tip: Clearly differentiate that Coulomb force acts between stationary charges and depends on distance and medium, while Lorentz force acts on moving charges in a magnetic field and has different dependencies.

 

Question 24. What is the Meissner effect?
Answer: The Meissner effect is a phenomenon where a superconductor, during its transition to the superconducting state, completely expels all magnetic flux from its interior. This means that a superconductor acts as a perfect diamagnet, preventing any magnetic field lines from penetrating it. This is a defining characteristic of superconductivity.
In simple words: The Meissner effect is when a superconductor pushes out all magnetic fields from inside itself as it becomes superconducting. It is like the material suddenly becomes perfectly non-magnetic on the inside.

🎯 Exam Tip: Define the Meissner effect as the complete expulsion of magnetic flux from a superconductor's interior during its transition to the superconducting state.

 

Question 25. Difference between Tangent Galvanometer and Moving coil galvanometer.
Answer:

In simple words: The Tangent Galvanometer uses the tangent law to measure magnetic fields and has a circular frame. The Moving Coil Galvanometer uses the force on a current coil in a magnetic field to measure voltage and has a rectangular coil.

🎯 Exam Tip: Highlight the different principles of operation (Tangent Law vs. torque on a current-carrying coil) and their primary uses (magnetic induction vs. potential/current measurement).

 

Question 26. Difference between the natural magnet and artificial magnet.
Answer:

In simple words: Natural magnets are found in nature, have irregular shapes, and include substances like lodestone. Artificial magnets are made by humans, can have specific shapes, and examples are bar magnets.

🎯 Exam Tip: Focus on the origin (natural vs. man-made), shape, and provide clear examples for each type of magnet.

XI. Three Mark Questions:

 

Question 1. State Curie-Weiss law.
Answer: The Curie-Weiss law describes the magnetic susceptibility of materials above their Curie temperature. It states that the magnetic susceptibility \( \text{X}_{\text{m}} \) of a ferromagnetic material above its Curie temperature \( \text{T}_{\text{C}} \) is given by:
\( \text{X}_{\text{m}} = \frac{\text{C}}{\text{T} - \text{T}_{\text{C}}} \)
Here, C is the Curie constant, T is the absolute temperature, and \( \text{T}_{\text{C}} \) is the Curie temperature. This law indicates that susceptibility decreases as temperature increases beyond the Curie point, where the material becomes paramagnetic.
In simple words: The Curie-Weiss law tells us how easily a material can be magnetized when its temperature is higher than a special point called the Curie temperature. It says that the material's magnetic strength goes down as it gets hotter.

🎯 Exam Tip: State the formula accurately and define all terms, especially emphasizing that it applies to ferromagnetic materials *above* their Curie temperature, where they behave paramagnetically.

 

Question 2. Define in terms of the Current Sensitivity of a galvanometer.
Answer: Current sensitivity of a galvanometer is defined as the deflection produced per unit current flowing through it. It is a measure of how much the galvanometer deflects for a given current. Mathematically, if \( \theta \) is the deflection and I is the current, then current sensitivity \( \text{I}_{\text{s}} = \frac{\theta}{\text{I}} = \frac{\text{NAB}}{\text{K}} \implies \text{I}_{\text{s}} = \frac{1}{\text{G}} \), where N is the number of turns, A is the area, B is the magnetic field, and K is the torsional constant. A higher current sensitivity means a larger deflection for a small current.
In simple words: Current sensitivity tells us how much a galvanometer's needle moves for each unit of electric current passing through it. A high sensitivity means it moves a lot even for a tiny current.

🎯 Exam Tip: Define current sensitivity as deflection per unit current, provide its formula \( (\text{I}_{\text{s}} = \theta/\text{I}) \), and explain that a higher value indicates greater ability to detect small currents.

 

Question 3. What are the applications of hysteresis loop?
Answer: The hysteresis loop is a valuable tool with several applications in magnetism:
1. It provides information about magnetic properties: The loop reveals key properties such as retentivity (magnetism retained after field removal), coercivity (field needed to demagnetize), permeability, susceptibility, and energy loss in a material. This helps us understand how a material responds to magnetic fields.
2. It helps in selecting suitable materials: By analyzing the shape and size of the hysteresis loop, engineers can choose the right material for specific purposes, such as permanent magnets (large loop area, high retentivity, high coercivity) or transformer cores (small loop area, low retentivity, low coercivity).
In simple words: The hysteresis loop helps us understand how magnetic materials behave. It shows how much magnetism a material holds onto and how much energy it loses, which helps in picking the right material for things like strong magnets or efficient transformers.

🎯 Exam Tip: Highlight the two main applications: characterizing magnetic material properties and guiding material selection for specific uses like permanent magnets or electromagnets/transformer cores.

 

Question 4. State Maxwell’s right-hand corkscrew rule.
Answer: Maxwell’s right-hand corkscrew rule is a convention used to determine the direction of a magnetic field produced by an electric current, or vice versa. The rule states:
1. If a right-handed corkscrew is rotated so that it advances in the direction of the current, then the direction of rotation of the screw gives the direction of the magnetic field lines.
2. Conversely, if the magnetic field direction is known, rotating the corkscrew in the direction of the field will show the current's direction by how the screw advances.
In simple words: Imagine twisting a corkscrew. If the corkscrew moves forward in the direction of the electric current, then the way you turn the corkscrew shows the direction of the magnetic field.

🎯 Exam Tip: State the rule clearly, linking the advancement of the corkscrew to current direction and its rotation to the magnetic field direction. This rule is a visual aid for directionality.

 

Question 5. Write short notes on MRI (Magnetic Resonance Imaging).
Answer: Magnetic Resonance Imaging (MRI) is a non-invasive medical imaging technique used to create detailed images of organs and soft tissues within the body. Its key aspects are:
1. Diagnostic and monitoring tool: MRI helps diagnose and monitor various abnormal conditions in the head, chest, abdomen, and pelvis, such as tumors, strokes, and spinal injuries.
2. Strong magnetic field: It uses a superconducting wire to produce a very strong magnetic field, which aligns the protons in the body's water molecules.
3. Radiofrequency pulses and computer: MRI uses radiofrequency pulses to temporarily knock these aligned protons out of alignment. When they realign, they emit signals that a computer processes to generate detailed images.
In simple words: MRI is a medical test that uses strong magnets and radio waves to take clear pictures of soft parts inside the body, like organs and muscles. It helps doctors find problems without surgery.

🎯 Exam Tip: Mention MRI's use in diagnostics, the generation of strong magnetic fields (superconducting magnets), and the role of radiofrequency pulses in creating detailed images of soft tissues.

 

Question 6. Difference between Coulomb’s law and Biot-Savart’s law.
Answer:

In simple words: Coulomb's law describes electric fields from charges and points straight out. Biot-Savart's law describes magnetic fields from current elements and is always at a right angle to both the current and the position.

🎯 Exam Tip: Emphasize the scalar nature of charge for Coulomb's law versus the vector nature of current elements for Biot-Savart's law, and the differing directional dependencies of the fields they describe.

 

Question 7. List out the deflection based on the velocity – velocity selector.
Answer: