GSEB Class 12 Physics Solutions Chapter 5 Magnetism and Matter

GSEB Solutions Class 12 Physics Chapter 5 Magnetism And Matter

 

Question 1. Answer the following questions regarding earth's magnetism.
(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth's magnetic field.
(b) The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain?
(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?
(d) In which direction would a compass free to move in the vertical plane point to, if on the geomagnetic north or south pole?
(e) The earth's field, it is claimed, roughly approximates the field due to a dipole of magnetic moment \(8 \times 10 \, JT^{-1}\) located at its centre. Check the order of magnitude of this number in some way.
(f) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth's surface oriented in different directions. How is such a thing possible at all?
Answer:
(a) The three physical quantities commonly used to define the Earth's magnetic field are:
- Magnetic declination
- Angle of dip
- The horizontal component of the Earth's magnetic field.
(b) Britain is located closer to the magnetic north pole. Therefore, the angle of dip in Britain is expected to be larger than in India, approximately 70°.
(c) In Melbourne, Australia, which is in the Southern Hemisphere, the Earth's magnetic field lines would appear to emerge from the ground.
(d) A magnetic compass, when free to move vertically at the geomagnetic north or south pole, can point in any direction. This is because the horizontal component of the Earth's magnetic field is zero at these specific points.
(e) To verify the order of magnitude, we can calculate the magnetic field \(B\) at the Earth's surface using the formula for a dipole:
\[ B = \frac{\mu_0 M}{4 \pi r^3} \]
Here, \( M = 8 \times 10^{22} \, JT^{-1} \) (as per the provided calculation, assuming this is the corrected value for \(M\)) and the Earth's radius \( r = 6.4 \times 10^6 \, m \).
Plugging in the values, we get \( B \approx 0.3 \, G \), which roughly matches the magnitude of the horizontal field at the Earth's surface.
(f) The Earth's magnetic field is not perfectly like a simple dipole. The presence of local N-S poles, oriented in various directions, is possible due to the differing compositions and deposits of magnetized minerals within the Earth's crust.
In simple words: The Earth's magnetic field is described by three main parts: how far east or west it points, how much it dips into the ground, and its strength across the ground. The field changes in different places and is caused by hot, moving iron inside the Earth. Sometimes, small local magnetic spots appear because of certain rocks.

🎯 Exam Tip: When describing Earth's magnetic field, remember to include all three specifying quantities: declination, dip, and horizontal component, as these are key for a complete understanding and good scores.

 

Question 2. Answer the following questions.
(a) The earth's magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?
(b) The earth's core is known to contain iron. Yet geologists do not regard this as a source of the earth's magnetism. Why?
(c) The charged currents in the outer conducting regions of the earth's core are thought to be responsible for earth's magnetism. What might be the 'battery' (i.e., the source of energy) to sustain these currents?
(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth's field in such distant past?
(e) The earth's field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?
(f) Interstellar space has an extremely weak magnetic field of the order of \(10^{-12} \, T\). Can such a weak field be of any significant consequence? Explain.
Answer:
(a) Yes, the Earth's magnetic field does change over time. Small changes can be observed yearly or over a few years. However, a noticeable and significant change in the field's strength or direction might take several hundred years.
(b) The Earth's core contains molten iron. Although iron is usually a ferromagnetic material, at the very high temperatures found in the Earth's core, it loses its ferromagnetic properties and becomes paramagnetic. Therefore, geologists do not consider it a direct source of the Earth's permanent magnetism.
(c) The exact source of energy, often called the 'battery', that powers the charged currents in the Earth's outer core is not fully understood. However, it is thought to be possibly related to the radioactivity present in the Earth's interior core, which generates heat and drives convection currents.
(d) Geologists can learn about the Earth's magnetic field from the very distant past (4 to 5 billion years ago) by studying the magnetic properties of ancient rocks. As molten rock cools and solidifies, the magnetic minerals within it align with the Earth's magnetic field at that time, essentially "locking in" a record of the field's direction and strength.
(e) At large distances, typically beyond 30,000 km, the Earth's magnetic field deviates significantly from a simple dipole shape. This distortion is mainly caused by the interaction of the Earth's magnetic field with the solar wind and charged particles in the ionosphere.
(f) Even a very weak magnetic field, like the \(10^{-12} \, T\) field in interstellar space, can have important cosmological effects. For example, newly formed particles in space can be accelerated to very high energies by this weak field over vast distances.
In simple words: The Earth's magnetic field changes slowly over time, and its source is the moving hot iron deep inside. We know about past changes from old rocks. Far from Earth, the field gets twisted by the sun's particles. Even very weak magnetic fields in space can push tiny particles a lot over long distances.

🎯 Exam Tip: For magnetism questions, remember to clarify "molten" iron versus "solid" iron in the core and how rock magnetization helps study Earth's ancient field. Mentioning solar wind and ionosphere for field distortion is also crucial for higher marks.

 

Question 3. A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to \(4.5 \times 10^{-2} \, J\). What is the magnitude of the magnetic moment of the magnet?
Answer:
Given values are:
Angle, \( \theta = 30^\circ \)
Magnetic field strength, \( B = 0.25 \, T \)
Torque, \( \tau = 4.5 \times 10^{-2} \, J \)
We need to find the magnetic moment, \( m \).
The formula for torque experienced by a magnet in a magnetic field is:
\( \tau = m B \sin \theta \)
To find \( m \), we rearrange the formula:
\( m = \frac{\tau}{B \sin \theta} \)
Substitute the given values:
\( m = \frac{4.5 \times 10^{-2} \, J}{0.25 \, T \times \sin 30^\circ} \)
Since \( \sin 30^\circ = 0.5 \):
\( m = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5} = \frac{4.5 \times 10^{-2}}{0.125} \)
\( m = 36 \times 10^{-2} = 0.36 \, Am^2 \)
The magnetic moment is \( 0.36 \, Am^2 \) (or \( JT^{-1} \)).
In simple words: A small magnet spinning in a magnetic field feels a twist. If we know how much it twists, how strong the field is, and the angle, we can find the magnet's strength, called its magnetic moment. Here, it is 0.36 Am².

🎯 Exam Tip: Remember the formula for torque on a magnetic dipole in a magnetic field: \( \tau = m B \sin \theta \). Pay attention to units and basic trigonometric values for accurate calculations.

 

Question 4. A short bar magnet of magnetic moment \(m = 0.32 \, JT^{-1}\) is placed in a uniform magnetic field of 0.15 T. In which orientation would it correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Answer:
Given values:
Magnetic moment, \( m = 0.32 \, JT^{-1} \)
Magnetic field, \( B = 0.15 \, T \)
The potential energy \( U \) of a magnetic dipole in a magnetic field is given by \( U = -mB \cos \theta \), where \( \theta \) is the angle between the magnetic moment vector and the magnetic field vector.

(a) For stable equilibrium, the magnetic moment of the magnet should be aligned parallel to the external magnetic field. In this case, \( \theta = 0^\circ \), so \( \cos 0^\circ = 1 \).
Potential energy \( U_{stable} = -mB \cos 0^\circ = -mB \)
\( U_{stable} = -0.32 \times 0.15 = -0.048 \, J \)
This can also be written as \( -4.8 \times 10^{-2} \, J \).

(b) For unstable equilibrium, the magnetic moment of the magnet should be aligned anti-parallel to the external magnetic field. In this case, \( \theta = 180^\circ \), so \( \cos 180^\circ = -1 \).
Potential energy \( U_{unstable} = -mB \cos 180^\circ = -mB (-1) = mB \)
\( U_{unstable} = 0.32 \times 0.15 = 0.048 \, J \)
This can also be written as \( 4.8 \times 10^{-2} \, J \).
In simple words: A magnet in a field wants to align itself. If it points the same way as the field, it's stable and has the lowest energy (negative energy). If it points the opposite way, it's unstable and has the highest energy (positive energy).

🎯 Exam Tip: Remember that stable equilibrium corresponds to the lowest potential energy when `\(\theta = 0^\circ\)`, and unstable equilibrium to the highest potential energy when `\(\theta = 180^\circ\)`. Always include units for energy calculations.

 

Question 5. A closely wound solenoid of 800 turns and area of cross-section \(2.5 \times 10^{-4} \, m^2\) carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Answer:
Given values are:
Number of turns, \( N = 800 \)
Area of cross-section, \( A = 2.5 \times 10^{-4} \, m^2 \)
Current, \( I = 3.0 \, A \)

A solenoid acts like a bar magnet because the magnetic field lines it produces are very similar to those of a bar magnet. Inside a long solenoid, the magnetic field is uniform and parallel to its axis, much like the field inside a bar magnet. Outside, the field lines emerge from one end and enter the other, defining a north pole and a south pole depending on the direction of the current.

The magnetic moment \( m \) of a solenoid is given by the formula:
\( m = N I A \)
Substitute the given values:
\( m = 800 \times 3.0 \, A \times 2.5 \times 10^{-4} \, m^2 \)
\( m = 2400 \times 2.5 \times 10^{-4} \)
\( m = 6000 \times 10^{-4} = 0.600 \, Am^2 \)
The associated magnetic moment is \( 0.600 \, Am^2 \).
In simple words: A coil of wire with current flowing through it acts like a simple bar magnet, creating similar magnetic fields. The strength of this "solenoid magnet" depends on how many turns it has, the current flowing through it, and the area of each turn. Here, its magnetic strength is 0.6 Am².

🎯 Exam Tip: Remember that a solenoid's magnetic field resembles a bar magnet's, with polarity determined by current direction. The magnetic moment formula \( m = NIA \) is fundamental for calculations.

 

Question 6. If the solenoid in Exercise. 5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Answer:
From Exercise 5, the magnetic moment of the solenoid is \( M = 0.6 \, Am^2 \).
Given values are:
Magnetic field, \( B = 0.25 \, T \)
Angle between the solenoid's axis and the magnetic field, \( \theta = 30^\circ \)
The formula for torque \( \tau \) on a magnetic dipole in a magnetic field is:
\( \tau = M B \sin \theta \)
Substitute the known values:
\( \tau = 0.6 \, Am^2 \times 0.25 \, T \times \sin 30^\circ \)
Since \( \sin 30^\circ = 0.5 \):
\( \tau = 0.6 \times 0.25 \times 0.5 \)
\( \tau = 0.075 \, Nm \)
The magnitude of the torque on the solenoid is \( 0.075 \, Nm \).
In simple words: When the solenoid from the last problem is placed in a new magnetic field at an angle, it will feel a twisting force (torque). The amount of twist depends on the solenoid's magnetic strength, the new field's strength, and the angle between them. Here, the twist is 0.075 Nm.

🎯 Exam Tip: Ensure you use the magnetic moment derived from the previous problem (if applicable) and correctly apply the torque formula \( \tau = MB \sin \theta \). Accurate trigonometric values for common angles are essential.

 

Question 7. A bar magnet of magnetic moment \(1.5 \, JT^{-1}\) lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?
Answer:
Given values:
Magnetic moment, \( m = 1.5 \, JT^{-1} \)
Magnetic field, \( B = 0.22 \, T \)
Initially, the magnet is aligned with the field, so the initial angle \( \theta_1 = 0^\circ \). The initial potential energy is \( U_1 = -mB \cos 0^\circ = -mB \).

(a) Work required by an external torque is the change in potential energy, \( W = U_2 - U_1 \).
(i) To align the magnetic moment normal to the field direction:
The final angle \( \theta_2 = 90^\circ \). The final potential energy \( U_2 = -mB \cos 90^\circ = 0 \).
Work required \( W_i = U_2 - U_1 = 0 - (-mB) = mB \)
\( W_i = 1.5 \times 0.22 = 0.33 \, J \).

(ii) To align the magnetic moment opposite to the field direction:
The final angle \( \theta_2 = 180^\circ \). The final potential energy \( U_2 = -mB \cos 180^\circ = -mB (-1) = mB \).
Work required \( W_{ii} = U_2 - U_1 = mB - (-mB) = 2mB \)
\( W_{ii} = 2 \times 1.5 \times 0.22 = 0.66 \, J \).

(b) The torque \( \tau \) on the magnet is given by \( \tau = mB \sin \theta \).
(i) When the magnet is normal to the field direction (i.e., \( \theta = 90^\circ \)):
\( \tau_i = mB \sin 90^\circ = mB \)
\( \tau_i = 1.5 \times 0.22 = 0.33 \, Nm \). This torque tends to turn the magnet towards the direction of the field.

(ii) When the magnet is opposite to the field direction (i.e., \( \theta = 180^\circ \)):
\( \tau_{ii} = mB \sin 180^\circ = 0 \)
The torque is zero in this orientation.
In simple words: Moving a magnet away from its aligned position in a magnetic field requires work. Turning it to be sideways costs some energy, and turning it to point exactly opposite costs twice that much. When it's sideways, it feels a strong twist. When it points opposite, it feels no twist, but it's not stable.

🎯 Exam Tip: Differentiate between potential energy and work done; work is the change in potential energy. Remember that torque is maximum when `\(\theta = 90^\circ\)` and zero when `\(\theta = 0^\circ\)` or `\(\theta = 180^\circ\)`. Always state the direction of torque if asked.

 

Question 8. A closely wound solenoid of 2000 turns and area of cross-section \(1.6 \times 10^{-4} \, m^2\), carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of \(7.5 \times 10^{-2} \, T\) is set up at an angle of 30° with the axis of the solenoid?
Answer:
Given values:
Number of turns, \( N = 2000 \)
Area of cross-section, \( A = 1.6 \times 10^{-4} \, m^2 \)
Current, \( I = 4.0 \, A \)

(a) The magnetic moment \( M \) associated with the solenoid is calculated using the formula:
\( M = N I A \)
Substitute the given values:
\( M = 2000 \times 4.0 \, A \times 1.6 \times 10^{-4} \, m^2 \)
\( M = 8000 \times 1.6 \times 10^{-4} \)
\( M = 12800 \times 10^{-4} = 1.28 \, Am^2 \)

(b) If a uniform horizontal magnetic field \( B = 7.5 \times 10^{-2} \, T \) is set up at an angle \( \theta = 30^\circ \) with the axis of the solenoid:
The net force on a solenoid in a *uniform* magnetic field is always zero.
The torque \( \tau \) on the solenoid is given by the formula:
\( \tau = M B \sin \theta \)
Substitute the calculated magnetic moment and given field values:
\( \tau = 1.28 \, Am^2 \times 7.5 \times 10^{-2} \, T \times \sin 30^\circ \)
Since \( \sin 30^\circ = 0.5 \):
\( \tau = 1.28 \times 7.5 \times 10^{-2} \times 0.5 \)
\( \tau = 0.048 \, Nm \)
In simple words: This problem asks us to find the magnetic strength of a coil and then how it behaves in another magnetic field. First, the coil's magnetic strength is found using its turns, current, and area. Then, in a steady magnetic field, the coil doesn't get pushed, but it does get twisted if it's at an angle.

🎯 Exam Tip: Remember that a solenoid in a *uniform* magnetic field experiences zero net force but a non-zero torque (unless `\(\theta = 0^\circ\)` or `\(\theta = 180^\circ\)`). Clearly state both force and torque for completeness.

 

Question 9. A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude \(5.0 \times 10^{-2} \, T\). The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of \(2.0 \, s^{-1}\). What is the moment of inertia of the coil about its axis of rotation?
Answer:
Given values:
Number of turns, \( N = 16 \)
Radius of the coil, \( r = 10 \, cm = 10 \times 10^{-2} \, m = 0.1 \, m \)
Current, \( I = 0.75 \, A \)
External magnetic field, \( B = 5.0 \times 10^{-2} \, T \)
Frequency of oscillation, \( \nu = 2.0 \, s^{-1} \)

First, calculate the magnetic moment \( m \) of the coil:
\( m = N I A = N I (\pi r^2) \)
\( m = 16 \times 0.75 \times \pi \times (0.1)^2 \)
\( m = 12 \pi \times 0.01 = 0.12\pi \, Am^2 \)

The frequency of oscillation \( \nu \) of a magnetic dipole in a magnetic field is given by:
\( \nu = \frac{1}{2\pi} \sqrt{\frac{mB}{I_{inertia}}} \)
Where \( I_{inertia} \) is the moment of inertia of the coil.
To find \( I_{inertia} \), we rearrange the formula:
\( 2\pi\nu = \sqrt{\frac{mB}{I_{inertia}}} \)
Squaring both sides:
\( (2\pi\nu)^2 = \frac{mB}{I_{inertia}} \)
\( I_{inertia} = \frac{mB}{(2\pi\nu)^2} \)
Substitute the calculated magnetic moment and other given values:
\( I_{inertia} = \frac{(0.12\pi \, Am^2) \times (5.0 \times 10^{-2} \, T)}{(2\pi \times 2.0 \, s^{-1})^2} \)
\( I_{inertia} = \frac{0.12\pi \times 5.0 \times 10^{-2}}{16\pi^2} \)
\( I_{inertia} = \frac{0.60 \times 10^{-2}}{16\pi} \)
Using \( \pi \approx \frac{22}{7} \):
\( I_{inertia} = \frac{0.006}{16 \times \frac{22}{7}} = \frac{0.006 \times 7}{16 \times 22} = \frac{0.042}{352} \)
\( I_{inertia} \approx 1.19 \times 10^{-4} \, kg \, m^2 \)
As per the solution's calculation in the text:
\( I = \frac{(10 \times 10^{-2})^2 \times 16 \times 0.75 \times 5.0 \times 10^{-2}}{4 \times \frac{22}{7} \times 4} \)
\( I = \frac{10^{-2} \times 16 \times 0.75 \times 5 \times 10^{-2} \times 7}{4 \times 4 \times 22} \)
\( I = \frac{0.75 \times 35 \times 10^{-4}}{22} \)
\( I \approx 1.2 \times 10^{-4} \, kg \, m^2 \)
The moment of inertia of the coil is approximately \( 1.2 \times 10^{-4} \, kg \, m^2 \).
In simple words: A coil carrying electricity behaves like a tiny magnet. When this coil is slightly pushed in another magnetic field, it swings back and forth. By measuring how fast it swings, we can figure out how hard it is to make it spin, which is called its moment of inertia. Here, it is about 1.2 x 10^-4 kg m².

🎯 Exam Tip: This question combines magnetic moment with oscillatory motion. Remember the formula for the frequency of oscillation of a magnetic dipole and be careful with algebraic rearrangement to solve for the moment of inertia. Ensure consistent units throughout the calculation.

 

Question 10. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth's magnetic field at the place.
Answer:
Given values:
Angle of dip, \( \theta = 22^\circ \)
Horizontal component of Earth's magnetic field, \( H_E = 0.35 \, G \)
We need to find the total magnitude of the Earth's magnetic field, \( B_E \).
The relationship between the horizontal component, total field, and angle of dip is:
\( H_E = B_E \cos \theta \)
Rearranging to find \( B_E \):
\( B_E = \frac{H_E}{\cos \theta} \)
Substitute the given values:
\( B_E = \frac{0.35 \, G}{\cos 22^\circ} \)
Using \( \cos 22^\circ \approx 0.927 \):
\( B_E = \frac{0.35}{0.927} \approx 0.377 \, G \)
Rounding to two significant figures as in the given answer:
\( B_E = 0.38 \, G \)
The magnitude of the Earth's magnetic field at that place is approximately \( 0.38 \, G \).
In simple words: A dipping compass needle shows how much the Earth's magnetic field points down. If we know the horizontal part of this field and the dip angle, we can calculate the total strength of the Earth's magnetic field. Here, it is 0.38 Gauss.

🎯 Exam Tip: Understand the relationship between the Earth's total magnetic field, its horizontal and vertical components, and the angle of dip. The formula \( H_E = B_E \cos \theta \) is key for calculations involving these quantities.

 

Question 11. At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth's field is measured to be 0.16 G. Specify the direction and magnitude of the earth's field at the location.
Answer:
Given values:
Magnetic declination, \( \phi = 12^\circ \) west of geographic north.
Angle of dip, \( \theta = 60^\circ \) (the north tip points 60° above the horizontal, meaning the field dips 60° below the horizontal for a standard north-seeking pole).
Horizontal component of Earth's magnetic field, \( H_E = 0.16 \, G \).

First, determine the magnitude of the Earth's magnetic field, \( B_E \).
The relationship is \( H_E = B_E \cos \theta \).
So, \( B_E = \frac{H_E}{\cos \theta} \)
Substitute the values:
\( B_E = \frac{0.16 \, G}{\cos 60^\circ} \)
Since \( \cos 60^\circ = 0.5 \):
\( B_E = \frac{0.16}{0.5} = 0.32 \, G \)

Now, specify the direction:
The Earth's magnetic field at this location has a magnitude of \( 0.32 \, G \). Its direction is \( 12^\circ \) west from the geographic north, and it dips \( 60^\circ \) below the horizontal in the magnetic meridian.
In simple words: At this place in Africa, the compass points a little to the west of true north, and a special needle dips 60 degrees. Knowing the horizontal strength, we find the total Earth's magnetic field is 0.32 Gauss. Its direction is 12 degrees west from true north.

🎯 Exam Tip: When asked to "specify the direction and magnitude," ensure you provide both components clearly. Magnetic declination defines the horizontal direction, and the angle of dip defines the vertical component relative to the horizontal. Don't forget units.

 

Question 12. A short bar magnet has a magnetic moment of \(0.48 \, JT^{-1}\). Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Answer:
Given values:
Magnetic moment, \( m = 0.48 \, JT^{-1} \)
Distance, \( r = 10 \, cm = 0.1 \, m \)
The constant \( \frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A \).

(a) Magnetic field on the axial line:
The formula for the magnetic field \( B_{axial} \) at a distance \( r \) from the centre of a short bar magnet on its axis is:
\[ B_{axial} = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \]
Substitute the values:
\( B_{axial} = 10^{-7} \times \frac{2 \times 0.48}{(0.1)^3} \)
\( B_{axial} = 10^{-7} \times \frac{0.96}{0.001} \)
\( B_{axial} = 0.96 \times 10^{-4} \, T \)
The direction of the magnetic field on the axial line is along the direction of the magnetic moment `\(\vec{m}\)`.

(b) Magnetic field on the equatorial line (normal bisector):
The formula for the magnetic field \( B_{equatorial} \) at a distance \( r \) from the centre of a short bar magnet on its equatorial line is:
\[ B_{equatorial} = \frac{\mu_0}{4\pi} \frac{m}{r^3} \]
Substitute the values:
\( B_{equatorial} = 10^{-7} \times \frac{0.48}{(0.1)^3} \)
\( B_{equatorial} = 10^{-7} \times \frac{0.48}{0.001} \)
\( B_{equatorial} = 0.48 \times 10^{-4} \, T \)
The direction of the magnetic field on the equatorial line is opposite to the direction of the magnetic moment `\(\vec{m}\)`.
In simple words: For a small magnet, the magnetic field is stronger along its length (axis) and points in the same direction as the magnet's strength. The field is weaker on its side (equatorial line) and points in the opposite direction. Here, at 10 cm, the axial field is \(0.96 \times 10^{-4} \, T\) and the equatorial field is \(0.48 \times 10^{-4} \, T\).

🎯 Exam Tip: Clearly distinguish between the axial and equatorial field formulas for a bar magnet. Remember that the axial field is twice the equatorial field at the same distance, and their directions relative to the magnetic moment are opposite.

 

Question 13. A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the center of the magnet. The earth's magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null point (i.e., 14 cm) from the centre of the magnet? (At null points, the field due to a magnet is equal and opposite to the horizontal component of the earth's magnetic field.)
Answer:
Given values:
Distance to null points, \( d = 14 \, cm = 0.14 \, m \)
Earth's magnetic field, \( B_E = 0.36 \, G = 0.36 \times 10^{-4} \, T \)
Angle of dip is zero, which means the Earth's horizontal component \( H_E = B_E = 0.36 \, G \).

At null points on the axis of the magnet, the magnet's field is equal and opposite to the Earth's horizontal field. So, \( B_{axial} = H_E \).
The formula for the axial magnetic field is \( B_{axial} = \frac{\mu_0}{4\pi} \frac{2m}{d^3} \).
Therefore, \( \frac{\mu_0}{4\pi} \frac{2m}{d^3} = H_E \).
Since \( \frac{\mu_0}{4\pi} = 10^{-7} \):
\( 10^{-7} \frac{2m}{(0.14)^3} = 0.36 \times 10^{-4} \)
From this, we can find \( m \). Alternatively, we recognize that on the equatorial line, the magnetic field \( B_{equatorial} \) is half of the axial field at the same distance:
\( B_{equatorial} = \frac{1}{2} B_{axial} = \frac{1}{2} H_E \)
So, \( B_{equatorial} = \frac{1}{2} \times 0.36 \, G = 0.18 \, G \).

When null points occur on the axis, it implies that the magnet's magnetic moment is oriented such that its axial field opposes the Earth's field. If the magnet's north pole points to the geographic north, the Earth's field and the magnet's axial field would add up (no null points). Thus, the magnet's north pole must point towards the geographic south, meaning its magnetic moment `\(\vec{m}\)` points geographic south.
On the normal bisector (equatorial line), the magnet's field direction is opposite to its magnetic moment. So, if `\(\vec{m}\)` points geographic south, `\(\vec{B}_{equatorial}\)` points geographic north.
Since the Earth's magnetic field \( H_E \) also points geographic north, the two fields on the equatorial line will add up.

Total magnetic field on the normal bisector at distance \( d \):
\( B_{total} = B_{equatorial} + H_E \)
\( B_{total} = 0.18 \, G + 0.36 \, G = 0.54 \, G \)
In simple words: We first understand how the magnet is placed to create "null points" where its field cancels the Earth's field. Knowing this, we can find the magnet's strength. Then, we calculate the magnet's field on its side (equatorial line) and add it to the Earth's field, because in this setup, they point in the same direction. The total field is 0.54 Gauss.

🎯 Exam Tip: For null point problems, determine the magnet's orientation relative to the Earth's field. Remember that the equatorial field is half the axial field at the same distance, and carefully consider vector addition or subtraction of fields based on their directions.

 

Question 14. If the bar magnet in exercise. 13 is turned around by 180°, where will the new null points be located?
Answer:
In Exercise 13, the magnet's axis was along the magnetic north-south direction, and null points were on its axis. This implies the magnet's magnetic moment `\(\vec{m}\)` was pointing opposite to the Earth's horizontal field `\(\vec{H}_E\)` (e.g., magnet's N pole pointing geographic south if Earth's field is geographic north).

If the bar magnet is turned around by 180°, its magnetic moment `\(\vec{m}\)` will now point in the same direction as `\(\vec{H}_E\)` (e.g., magnet's N pole pointing geographic north).
In this new orientation:
- On the axial line, the magnet's field will add to the Earth's field, so no null points can be found there.
- On the equatorial line, the magnet's field `\(\vec{B}_{equatorial}\)` will be opposite to its magnetic moment `\(\vec{m}\)`. Since `\(\vec{m}\)` is now in the direction of `\(\vec{H}_E\)` (geographic north), `\(\vec{B}_{equatorial}\)` will point geographic south. This allows `\(\vec{B}_{equatorial}\)` to oppose `\(\vec{H}_E\)` (geographic north), creating null points on the equatorial line.

At a null point on the equatorial line, the magnitude of the magnet's field must equal the Earth's horizontal field:
\( B_{equatorial} = H_E \)
\[ \frac{\mu_0}{4\pi} \frac{m}{(d')^3} = H_E \]
From Exercise 13, we had \( \frac{\mu_0}{4\pi} \frac{2m}{d^3} = H_E \) for null points on the axis, where \( d = 14 \, cm \).
Equating the expressions for \( H_E \):
\( \frac{\mu_0}{4\pi} \frac{m}{(d')^3} = \frac{\mu_0}{4\pi} \frac{2m}{d^3} \)
\( \frac{1}{(d')^3} = \frac{2}{d^3} \)
\( (d')^3 = \frac{d^3}{2} \)
\( d' = d \left( \frac{1}{2} \right)^{1/3} \)
Substitute \( d = 14 \, cm \):
\( d' = 14 \times (0.5)^{1/3} \)
\( d' = 14 \times 0.7937 \approx 11.11 \, cm \)
The new null points will be located on the equatorial line at a distance of approximately \( 11.1 \, cm \) from the center of the magnet.
In simple words: When the magnet is flipped, the null points move. Instead of being along the magnet's main line, they will now be on its side (equatorial line). This is because the magnetic field directions change when the magnet is turned, causing it to cancel out Earth's field at these new spots. The new distance is about 11.1 cm.

🎯 Exam Tip: For problems involving magnet orientation and null points, always visualize or sketch the field lines to determine where cancellation can occur. Remember that the field on the equatorial line is related to the axial field by a factor of two, which simplifies distance calculations.

 

Question 15. A short bar magnet of magnetic moment \(5.25 \times 10^{-2} \, JT^{-1}\) is placed with its axis perpendicular to the earth's field direction. At what distance from the centre of the magnet is the resultant field inclined at 45° with earth's field on (a) its normal bisector and (b) its axis. Magnitude of the earth's field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Answer:
Given values:
Magnetic moment, \( m = 5.25 \times 10^{-2} \, JT^{-1} \)
Earth's magnetic field, \( B_E = 0.42 \, G = 0.42 \times 10^{-4} \, T \)
The magnet's axis is perpendicular to the Earth's field `\(\vec{B}_E\)`. This means the magnet's magnetic moment `\(\vec{m}\)` is perpendicular to `\(\vec{B}_E\)`.
The resultant magnetic field is inclined at \( 45^\circ \) with the Earth's field. This implies that the magnitude of the magnet's field (`\(B_{magnet}\)`) at that point must be equal to the magnitude of the Earth's field (`\(B_E\)`), because \( \tan 45^\circ = \frac{B_{magnet}}{B_E} = 1 \).
So, \( B_{magnet} = B_E = 0.42 \times 10^{-4} \, T \).

(a) On its normal bisector (equatorial line):
The magnetic field on the equatorial line \( B_{equatorial} \) is given by:
\[ B_{equatorial} = \frac{\mu_0}{4\pi} \frac{m}{d^3} \]
We set \( B_{equatorial} = B_E \):
\( 10^{-7} \frac{m}{d^3} = B_E \)
\( d^3 = \frac{10^{-7} \times m}{B_E} \)
Substitute the values:
\( d^3 = \frac{10^{-7} \times 5.25 \times 10^{-2}}{0.42 \times 10^{-4}} \)
\( d^3 = \frac{5.25 \times 10^{-9}}{0.42 \times 10^{-4}} = \frac{5.25}{0.42} \times 10^{-5} = 12.5 \times 10^{-5} = 0.000125 \, m^3 \)
\( d = (0.000125)^{1/3} = 0.05 \, m = 5.0 \, cm \).

(b) On its axis:
The magnetic field on the axial line \( B_{axial} \) is given by:
\[ B_{axial} = \frac{\mu_0}{4\pi} \frac{2m}{d^3} \]
We set \( B_{axial} = B_E \):
\( 10^{-7} \frac{2m}{d^3} = B_E \)
\( d^3 = \frac{10^{-7} \times 2m}{B_E} \)
Substitute the values:
\( d^3 = \frac{10^{-7} \times 2 \times 5.25 \times 10^{-2}}{0.42 \times 10^{-4}} \)
\( d^3 = \frac{10.5 \times 10^{-9}}{0.42 \times 10^{-4}} = \frac{10.5}{0.42} \times 10^{-5} = 25 \times 10^{-5} = 0.00025 \, m^3 \)
\( d = (0.00025)^{1/3} \approx 0.063 \, m = 6.3 \, cm \).
In simple words: A small magnet is placed sideways to the Earth's magnetic field. We want to find where its own field, when combined with Earth's field, points at a 45-degree angle. This happens when the magnet's field is as strong as Earth's field. We calculate the required distance for both the side and the end of the magnet. On its side, the distance is 5.0 cm, and on its end, it is 6.3 cm.

🎯 Exam Tip: The condition for the resultant field to be at 45° implies that the magnet's field component perpendicular to the Earth's field is equal in magnitude to the Earth's field. Be precise in applying axial versus equatorial field formulas and cuberoot calculations.

 

Question 16. Answer the following questions.
1. Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?
2. Why is diamagnetism, in contrast, almost independent of temperature?
3. If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
4. Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?
5. Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?
6. Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?
Answer:
1. When a paramagnetic sample is cooled, the random thermal motion of its atomic dipoles decreases. This allows a greater number of these dipoles to align with the external magnetizing field, resulting in a stronger overall magnetization for the same applied field.
2. Diamagnetism is a property that arises from the induced orbital motion of electrons in atoms, which opposes the external magnetic field (Lenz's Law). This induced effect is not significantly influenced by temperature changes because it's an intrinsic atomic response, unlike the alignment of permanent dipoles in paramagnets.
3. Bismuth is a diamagnetic material. Diamagnetic materials slightly repel external magnetic fields. Therefore, if a toroid uses bismuth as its core, the magnetic field inside the core will be slightly less than if the core were empty (i.e., filled with vacuum).
4. No, the permeability of a ferromagnetic material is not independent of the magnetic field. It shows a complex, non-linear relationship (hysteresis). Generally, the permeability is greater for lower magnetic fields before saturation is reached.
5. Ferromagnetic materials possess a very high relative permeability. This property causes magnetic field lines to preferentially enter or exit the material perpendicular to its surface. This behavior minimizes the magnetic reluctance (resistance) for the field lines, similar to how electric field lines enter or leave a conductor normally to minimize electrical resistance at the surface.
6. Yes, the maximum possible magnetization of a paramagnetic sample can indeed be of the same order of magnitude as that of a ferromagnet. However, this level of magnetization for a paramagnet is achievable only under very specific and extreme conditions, such as extremely strong magnetizing fields combined with very low temperatures.
In simple words: When things that are slightly magnetic (paramagnetic) get cold, their tiny magnets line up better with an outside field, making them more magnetic. Things that are weakly repelled by magnets (diamagnetic) don't care about temperature. Using bismuth in a magnetic coil makes the field inside slightly weaker. Strong magnets have permeability that changes with the field, being higher at lower fields. Magnetic lines always enter and leave strong magnets straight on. And yes, a weak magnet can become as strong as a strong one, but only if it's super cold and in a super strong field.

🎯 Exam Tip: For material properties, clearly define each type of magnetism (dia-, para-, ferro-), relate them to temperature dependence, and understand their effect on magnetic field lines. For paramagnetism, remember the dual condition of high field and low temperature for saturation.

Question 23. A sample of paramagnetic salt contains \(2.0 \times 10^{24}\) atomic dipoles, each with a dipole moment of \(1.5 \times 10^{-24} \text{ JT}^{-1}\). This sample is placed in a uniform magnetic field of 0.64 T and cooled to a temperature of 4.2 K. The magnetic saturation achieved is 15%. What is the total dipole moment of the sample when the magnetic field is 0.98 T and the temperature is 2.8 K? (Assume Curie's law)

Answer:

Under the first conditions, the total magnetic moment is calculated as:

\(1.5 \times 10^{-23} \times 2.0 \times 10^{24} \times \frac{15}{100} = 4.5 \text{ JT}^{-1}\)

According to Curie's law, magnetization (M) is directly proportional to the magnetic field (B) and inversely proportional to the temperature (T), so \(M = C \frac{B}{T}\).

Therefore, we can write: \(\frac{M'}{M} = \frac{B'}{T'} \times \frac{T}{B}\)

So, \(M' = M \times \frac{B'}{T'} \times \frac{T}{B}\)

Substituting the given values: \(M' = 4.5 \times \frac{0.98}{2.8} \times \frac{4.2}{0.64}\)

\(M' = 7.9 \text{ JT}^{-1}\)

In simple words: We used Curie's law to find how the total magnetic strength of the salt changes when the magnetic field and temperature are different. The law states that a material's magnetism depends on the field strength and temperature.

🎯 Exam Tip: Remember Curie's Law for paramagnetic materials, which describes how their magnetization changes with temperature and magnetic field. Be careful with unit conversions, especially for dipole moment and temperature. Ensure all calculations are correctly performed to find the final magnetic moment.

Question 24. A Rowland ring with a mean radius of 15 cm has 3500 turns of wire wound on a ferromagnetic core with a relative permeability of 800. What is the magnetic field B in the core for a magnetizing current of 1.2 A?

Answer:

Given values are: Radius, \(R = 15 \text{ cm} = 0.15 \text{ m}\); Number of turns, \(N = 3500\); Relative permeability, \(\mu_r = 800\); Current, \(I = 1.2 \text{ A}\).

The magnetic field B in the core of a Rowland ring is given by the formula:

\(B = \frac{\mu_0 \mu_r N I}{2 \pi R}\)

Where \(\mu_0\) is the permeability of free space, \(\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}\).

Substitute the values into the formula:

\(B = \frac{(4\pi \times 10^{-7}) \times 800 \times 3500 \times 1.2}{2 \pi \times 0.15}\)

Simplify the expression:

\(B = \frac{4 \times 10^{-7} \times 800 \times 3500 \times 1.2}{2 \times 0.15}\)

\(B = 4.48 \text{ T}\)

In simple words: We calculated the magnetic field inside a coil wrapped around a special iron core. We used a formula that considers how many times the wire is wound, how strong the current is, and how easily the core material allows magnetism to pass through it.

🎯 Exam Tip: For problems involving Rowland rings, remember the formula for magnetic field B inside the core, incorporating both absolute and relative permeability. Pay attention to unit conversions, especially for radius. Clearly identify all given parameters before substitution.

Question 25. The magnetic moment vectors \(\mu_s\) and \(\mu_l\) associated with the intrinsic spin angular momentum S and orbital angular momentum I, respectively, of an electron, are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by:

\(\mu_s = \frac{e}{m}S\)

\(\mu_l = \frac{e}{2m}I\)

Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

Answer:

The relation for \(\mu_l\) (orbital magnetic moment) is consistent with classical theory.

To derive this classically, consider an electron of mass 'm' and charge '-e' orbiting in a circular path of radius 'r' with a period 'T'.

The current (I) produced by this orbiting electron is \(I = \frac{e}{T}\).

The angular speed (\(\omega\)) is \(\omega = \frac{2\pi}{T}\), so \(T = \frac{2\pi}{\omega}\).

Substituting T, the current becomes \(I = \frac{e\omega}{2\pi}\).

The magnetic moment (\(\mu_l\)) for a current loop is given by \(IA\), where A is the area of the loop. For a circular orbit, \(A = \pi r^2\).

So, \(\mu_l = IA = \left(\frac{e\omega}{2\pi}\right) (\pi r^2) = \frac{1}{2} e \omega r^2\).

The orbital angular momentum (I) of the electron is \(I = mvr\). Since \(v = \omega r\), then \(I = m (\omega r) r = m \omega r^2\).

From this, \(\omega r^2 = \frac{I}{m}\).

Substitute \(\omega r^2\) back into the magnetic moment equation:

\(\mu_l = \frac{1}{2} e \left(\frac{I}{m}\right) = \frac{e}{2m}I\).

Since the electron's charge is negative, the magnetic moment and angular momentum are in opposite directions, and both are perpendicular to the plane of the orbit.

In contrast, the spin magnetic moment \(\mu_s = \frac{e}{m}S\) has a factor of \(\frac{e}{m}\) instead of \(\frac{e}{2m}\), which is twice the classical value and is an important result from quantum theory that cannot be explained classically.

In simple words: The formula for the magnetic strength from an electron's orbit matches what we expect from basic physics rules. We found this by imagining an electron moving in a circle, which creates a small electric current, and then calculating the magnetic effect of that current. The spin magnetic strength, however, is twice as large as expected from simple physics, showing a difference that quantum physics explains.

🎯 Exam Tip: Distinguish between orbital and spin magnetic moments. Be ready to classically derive the orbital magnetic moment starting from current definition and angular momentum. Remember that the spin magnetic moment cannot be derived classically and represents a fundamental quantum property. Pay attention to the relationship between magnetic moment, angular momentum, charge, and mass.

GSEB Class 12 Physics Magnetism And Matter Additional Important Questions And Answers

Question 1. Why does the magnet always align in a particular direction?

Answer:

A freely suspended magnet always points in a specific direction, which is roughly the geographical north-south. This alignment occurs because the Earth itself acts as a large magnet, exerting a force on the suspended magnet's poles.

In simple words: A free magnet aligns north-south because Earth acts like a giant magnet, pulling on it.

🎯 Exam Tip: Focus on understanding Earth's magnetic field and how it interacts with other magnets. This is a foundational concept in magnetism.

Question 2. What is meant by geographic north pole?

Answer:

The geographic north pole refers to the northernmost point on the Earth. It is the end point of the Earth's axis of rotation in the northern hemisphere, perpendicular to the equator's plane.

In simple words: The geographic north pole is the northernmost point on Earth, where the Earth's spinning axis comes out.

🎯 Exam Tip: Clearly differentiate between geographic poles (related to Earth's rotation) and magnetic poles (related to Earth's magnetic field). Accuracy in definitions is crucial.

Question 3. What is meant by geographic south pole?

Answer:

The geographic south pole is the southernmost point on Earth. It is the end point of the Earth's axis of rotation in the southern hemisphere, perpendicular to the equator's plane.

In simple words: The geographic south pole is the southernmost point on Earth, where the Earth's spinning axis comes out.

🎯 Exam Tip: Similar to the geographic north pole, ensure a clear definition. Understanding the Earth's rotational axis is key here.

Question 4. The north pole of a freely suspended magnet points towards the geographical north. Why?

Answer:

The north pole of a freely suspended magnet points towards the Earth's geographical north because the Earth's magnetic south pole is located near the geographical north pole. Opposite poles attract, causing the magnet to align this way.

In simple words: A magnet's north pole points to Earth's geographic north because Earth has a magnetic south pole there, and opposite poles attract.

🎯 Exam Tip: It's important to remember that Earth's magnetic south pole is near the geographic north pole, and vice versa. This explains why a compass points north.

Question 5. Does the magnetic south pole and geographic north pole coincide?

Answer:

No, the Earth's magnetic south pole and the geographic north pole do not coincide. They are close to each other but are not in the exact same location, leading to magnetic declination.

In simple words: No, the magnetic south pole of Earth is not exactly at the geographic north pole; they are slightly different.

🎯 Exam Tip: Understand that the Earth's magnetic poles are not fixed and do not perfectly align with the geographic poles. This difference is called magnetic declination.

Question 6. What is meant by geographic meridian and magnetic meridian?

Answer:

A geographic meridian is a vertical plane that passes through a specific location on Earth and contains the Earth's geographic axis (axis of rotation). A magnetic meridian, on the other hand, is a vertical plane that passes through a location and contains the direction of the Earth's magnetic field lines, connecting the magnetic north and south poles.

In simple words: A geographic meridian is a plane showing north-south based on Earth's spin axis. A magnetic meridian is a plane showing north-south based on Earth's magnetic field.

🎯 Exam Tip: Clearly define both meridians and emphasize the difference in their reference points: Earth's rotational axis for geographic, and Earth's magnetic field for magnetic.

Question 7. Do these two meridians coincide anywhere?

Answer:

Yes, these two meridians can coincide in certain places on Earth. When they do, the magnetic declination at that location is zero. Places where this property holds true are connected by imaginary lines known as agonic lines.

In simple words: Yes, sometimes these two lines match up, which means there is no magnetic declination. Such places are linked by "agonic lines."

🎯 Exam Tip: Understand the concept of "agonic lines" as locations where magnetic declination is zero, indicating the coincidence of geographic and magnetic meridians.

Question 8. When a compass needle is pivoted so that it can rotate in a vertical plane (dip needle), will it align horizontally?

Answer:

Generally, a dip needle will not align horizontally. When properly adjusted and allowed to move in a vertical plane, it will show an angle with the horizontal, which is called the angle of dip or inclination, reflecting the vertical component of Earth's magnetic field.

In simple words: Usually, a dip needle will not lie flat. It shows an angle with the ground, called the angle of dip, because Earth's magnetism pulls it up or down.

🎯 Exam Tip: Recognize the function of a dip needle and its relationship to the angle of dip. Understand that horizontal alignment only occurs at the magnetic equator.

Question 9. If it is not aligning horizontally, what does it mean?

Answer:

If a dip needle is not aligning horizontally, it indicates that the Earth's total magnetic field at that location has a vertical component. This means the net magnetic field is not purely horizontal.

In simple words: If a dip needle isn't flat, it means Earth's magnetic field has a part pulling it up or down, so the total field isn't just flat.

🎯 Exam Tip: Connect the non-horizontal alignment of a dip needle directly to the presence of a vertical component of Earth's magnetic field, signifying a non-zero angle of dip.

Question 10. If the above dip needle is taken to the pole and the magnetic equator, how it will align?

Answer:

At the magnetic poles, the angle of dip is 90°, so the dip needle will stand vertically. At the magnetic equator, the angle of dip is 0°, meaning the dip needle will align horizontally.

In simple words: At Earth's magnetic poles, the dip needle stands straight up. At the magnetic equator, it lies flat.

🎯 Exam Tip: Remember the extreme cases of dip angle: 90° at the magnetic poles (vertical alignment) and 0° at the magnetic equator (horizontal alignment).

Question 11. How can we resolve the earth's magnetic field at a place into two rectangular components?

Answer:

The Earth's magnetic field (\(B_E\)) at any point can be broken down into two components: a horizontal component (\(H_E\)) and a vertical component (\(Z_E\)). If \(\theta\) is the angle of dip at that place, then \(H_E = B_E \cos\theta\) and \(Z_E = B_E \sin\theta\).

In simple words: We can split Earth's total magnetic field into two parts: one that goes sideways (horizontal) and one that goes up or down (vertical). These parts depend on the "dip angle" of the magnetic field.

🎯 Exam Tip: Master the trigonometric relations for horizontal and vertical components of Earth's magnetic field, \(H_E = B_E \cos\theta\) and \(Z_E = B_E \sin\theta\), where \(\theta\) is the angle of dip. This is fundamental for vector decomposition.

Question 12. What are null points?

Answer:

Null points are specific locations where the net magnetic field intensity is zero. This occurs when the magnetic field produced by a magnet is exactly equal in magnitude and opposite in direction to the Earth's magnetic field at that point.

In simple words: Null points are spots where all magnetic fields, like from a magnet and Earth, cancel each other out, making the total magnetic force zero.

🎯 Exam Tip: Define null points as locations of zero net magnetic field. Emphasize that this cancellation happens when the magnet's field and Earth's field are equal and opposite.

Question 13. What are the magnetic fields along the axis and equatorial line of a bar magnet?

Answer:

For a short bar magnet, the magnetic field along its axial line at a distance 'd' from its center is \(B_{axial} = \frac{\mu_0}{4\pi} \frac{2Md}{(d^2 - l^2)^2}\), where M is the magnetic moment and '2l' is the length of the magnet. If the magnet is very short (\(l \ll d\)), then \(B_{axial} = \frac{\mu_0}{4\pi} \frac{2M}{d^3}\). The direction is along the magnetic moment vector.

The magnetic field along its equatorial line (normal bisector) at a distance 'd' from its center is \(B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{(d^2 + l^2)^{3/2}}\). If the magnet is very short (\(l \ll d\)), then \(B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{d^3}\). The direction is opposite to the magnetic moment vector.

In simple words: For a bar magnet, the magnetic field is strongest and points in the same direction as the magnet's axis (along the axial line). Along the middle line (equatorial line), the field is weaker and points in the opposite direction.

🎯 Exam Tip: Remember the formulas for magnetic field strength on the axial and equatorial lines of a bar magnet, especially the inverse cube dependence on distance for short magnets. Note the directional difference: same direction as magnetic moment on axial line, opposite on equatorial line.

Question 14. Null points are obtained on the axial line when the north pole of the bar magnet is pointing towards the geographical south. Why?

Answer:

When the north pole of a bar magnet points towards the geographical south, the magnetic field lines from the magnet along its axial line are directed from south to north. Since the Earth's magnetic field lines also point from geographical south to geographical north, the magnet's field and Earth's field become equal in magnitude and opposite in direction on the axial line, allowing for null points.

Null points cannot form on the equatorial line in this setup because the magnet's field lines there would be parallel to the Earth's field, not opposite.

In simple words: Null points appear on the axial line when the magnet's north pole faces geographic south because the magnet's field points one way (south to north) and Earth's field points the other (south to north). These fields can cancel each other out, but not on the equatorial line where they would be in the same direction.

🎯 Exam Tip: Visualize the direction of magnetic field lines from both the bar magnet and Earth to determine where cancellation (and thus null points) can occur. The key is understanding vector addition of magnetic fields.

Question 15. Null points are obtained on the equatorial line when the north pole of the bar magnet is pointing towards the geographical north. Why?

Answer:

When the north pole of a bar magnet points towards the geographical north, the magnetic field lines from the magnet along its equatorial line are directed opposite to the magnet's magnetic moment, which means they are pointing southwards (opposite to geographic north). Since the Earth's magnetic field lines point from geographical south to geographical north, on the equatorial line, the magnet's field and Earth's field are opposite in direction. If their magnitudes are equal, they cancel out, forming null points.

In simple words: Null points are found on the middle line of a magnet when its north pole faces geographic north. This happens because the magnet's field on its middle line points south, which is opposite to Earth's magnetic field (which points north). When these fields are equal and opposite, they cancel out.

🎯 Exam Tip: For null points on the equatorial line, remember that the magnet's field is opposite to its magnetic moment. This configuration is necessary for it to oppose and cancel Earth's magnetic field, which points towards the geographic north.

Question 16. A magnetic substance is placed in magnetic fields of different intensities. Does it produce the same effect?

Answer:

No, a magnetic substance does not produce the same effect when placed in magnetic fields of different intensities. The magnetization of the material depends directly on the strength of the external magnetizing field. A stronger external field generally results in a greater response or higher magnetization of the material.

In simple words: No, a magnetic material acts differently in strong versus weak magnetic fields. Its magnetic response changes with how strong the external field is.

🎯 Exam Tip: Understand that the induced magnetism in a material is generally proportional to the applied magnetizing field (H). This relationship is quantified by magnetic susceptibility and permeability, which vary for different materials and field strengths.

Question 17. After magnetizing a substance, by measuring which quantity we can understand the extent to which it is magnetized?

Answer:

To understand how much a substance has been magnetized, we measure its intensity of magnetization, denoted as \(\bar{M}\). This quantity is defined as the magnetic moment per unit volume of the material.

In simple words: After magnetizing something, we measure its "intensity of magnetization" to see how strongly it has become a magnet. This is the magnet's strength per amount of space it takes up.

🎯 Exam Tip: Know the definition of intensity of magnetization (\(M\)) as magnetic moment per unit volume. This is a key parameter for describing the magnetic state of a material.

Question 18. When we apply an external magnetizing field to a material, what will be the resultant magnetic field inside it?

Answer:

When an external magnetizing field is applied to a material, the resultant magnetic field inside it is the vector sum of the external field and the field produced by the magnetization of the material itself. This can be expressed as \(B = \mu_0 H \pm \mu_0 M\), where \(\mu_0 H\) is the magnetic induction due to the external field (H) and \(\mu_0 M\) is the magnetic induction due to the material's magnetization (M). The sign depends on whether the induced field adds to or subtracts from the external field.

In simple words: The total magnetic field inside a material is a mix of the field we apply from outside and the field the material makes inside itself. These fields can add up or subtract from each other.

🎯 Exam Tip: Remember that the total magnetic field (B) inside a material is a superposition of the applied magnetizing field (H) and the field due to the material's magnetization (M). Understand the relation \(B = \mu_0 (H+M)\) for most materials, or \(B = \mu_0 H \pm \mu_0 M\) considering diamagnetic and paramagnetic responses.

Question 19. Is there any relation between B and H?

Answer:

Yes, there is a direct relationship between magnetic induction (B) and magnetizing field strength (H). The ratio of magnetic induction to magnetizing field strength is defined as the magnetic permeability (\(\mu\)) of the material. So, \(\mu = \frac{B}{H}\). In free space, this is \(\mu_0\), and for materials, it is \(\mu = \mu_0 \mu_r\), where \(\mu_r\) is the relative permeability.

In simple words: Yes, magnetic induction (B) and magnetizing field strength (H) are related. How easily a material lets magnetism pass through it (its permeability) links them together.

🎯 Exam Tip: Define magnetic permeability (\(\mu\)) as the ratio \(\frac{B}{H}\). Understand its significance in describing how a material responds to a magnetizing field and differentiate between absolute permeability (\(\mu\)) and relative permeability (\(\mu_r\)).

Question 20. Is there any relation between M and H?

Answer:

Yes, there is a relationship between the intensity of magnetization (M) and the magnetizing field strength (H). The ratio of the intensity of magnetization to the magnetizing field strength is defined as the magnetic susceptibility (\(\chi\)) of the material. Therefore, \(\chi = \frac{M}{H}\).

In simple words: Yes, the strength of magnetism created inside a material (M) is related to how strong the outside magnetic field is (H). This relationship is called "magnetic susceptibility."

🎯 Exam Tip: Define magnetic susceptibility (\(\chi\)) as the ratio \(\frac{M}{H}\). This parameter indicates how easily a material can be magnetized when an external field is applied. Remember its role in distinguishing different types of magnetic materials.

Question 21. Do all materials have a net magnetic moment in the normal state?

Answer:

No, not all materials have a net magnetic moment in their normal, unmagnetized state. For many substances, especially diamagnetic ones, the individual atomic magnetic moments cancel out, resulting in a zero net magnetic moment.

In simple words: No, not all materials naturally have a magnetic strength. In many materials, the tiny internal magnets cancel each other out, leaving no overall magnetic strength.

🎯 Exam Tip: Understand that the presence or absence of a net magnetic moment in the normal state is a key differentiator for diamagnetic, paramagnetic, and ferromagnetic materials at an atomic level.

Question 22. What is the reason for the magnetic moment of a material?

Answer:

The magnetic moment of a material primarily originates from the orbital and spin properties of the electrons within its atoms. Electrons orbiting the nucleus act like tiny current loops, creating an orbital magnetic moment, and electrons themselves possess an intrinsic spin, which also contributes a spin magnetic moment.

In simple words: A material's magnetic strength comes from its electrons. Electrons spin and move around the atom's center, both of which create tiny magnetic effects.

🎯 Exam Tip: Recall that atomic magnetic moments arise from both the orbital motion of electrons (like tiny current loops) and their intrinsic spin. This is a fundamental concept in understanding the magnetic properties of matter.

Question 23. Do all materials behave in the same manner in a magnetic field?

Answer:

No, materials do not all behave in the same way when placed in a magnetic field. Different magnetic materials, such as paramagnetic, diamagnetic, and ferromagnetic substances, exhibit distinct responses to external magnetic fields due to their varying internal atomic structures and electron properties.

In simple words: No, different materials react differently to magnets. Some are weakly attracted, some are weakly pushed away, and some are strongly attracted.

🎯 Exam Tip: Be able to differentiate the behavior of diamagnetic, paramagnetic, and ferromagnetic materials in a magnetic field. This distinction is central to classifying magnetic substances.

Question 24. Pick the odd one out of the following based on torque.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र चार अलग-अलग स्थितियों को दिखाता है: (a) एक विद्युत क्षेत्र में आवेश, (b) एक चुंबक, (c) एक धारावाही लूप, और (d) एक विद्युत क्षेत्र में आवेश। इसमें से हमें उस वस्तु को पहचानना है जो चुंबकीय क्षेत्र में बल आघूर्ण के आधार पर बाकी से अलग है।

(a) (b) (c) (d)

Answer:

The odd one out is (a) and (d).

In simple words: The pictures showing a charge in an electric field (a) and (d) are different from the magnet (b) and current loop (c) in terms of how they would experience twisting forces (torque) in a magnetic field. Magnets and current loops feel magnetic torque, while single charges primarily feel magnetic force, not torque in the same way.

🎯 Exam Tip: Understand that torque in a magnetic field acts on magnetic dipoles (like bar magnets and current loops) or current-carrying wires, causing rotation. A single charge primarily experiences a Lorentz force, not a rotational torque in the same context.

Question 25. The interaction of electric charges \(q_1\) and \(q_2\), separated by a distance r, is \(F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}\). If \(p_1\) and \(p_2\) are the properties of two magnetic poles, separated by a distance r, the force is \(F = \frac{\mu_0}{4 \pi} \frac{p_1 p_2}{r^2}\).

(a) Write two similar properties of these two forces.

(b) Name the quantities \(p_1\), \(p_2\), and \(\mu_0\).

(c) Show that 'p' has the unit ampere-meter.

Answer:

(a) Two similar properties of these forces are:

(i) Both are inverse square law forces, meaning the force strength decreases with the square of the distance between them.
(ii) Both are long-range forces and can be either attractive or repulsive, depending on the nature of the interacting charges or poles.

(b) In the magnetic force formula:

\(p_1\) and \(p_2\) are called pole strengths.
\(\mu_0\) is the permeability of vacuum (or air) and is a measure of how easily a magnetic field can be established in a vacuum.

(c) To show that 'p' has the unit ampere-meter, let's analyze the magnetic force formula: \(F = \frac{\mu_0}{4 \pi} \frac{p_1 p_2}{r^2}\)

We know the unit of Force (F) is Newtons (N).
The unit of \(\mu_0\) is Henry per meter (H/m) or \((\text{T} \cdot \text{m}) / \text{A}\) or \((\text{N}) / \text{A}^2\). Let's use \(N/A^2\).

So, \(N = \frac{N}{A^2} \frac{p^2}{m^2}\)

\(p^2 = A^2 m^2\)

\(p = Am\) (Ampere-meter)

Alternatively, from the relation for magnetic moment \(M = p \times 2l\), where M has units of \(Am^2\) and length (2l) has units of m.

So, \(Am^2 = p \times m\)

\(p = Am\)

In simple words: (a) Both electric and magnetic forces get weaker with distance in the same way, and they can pull or push objects far away. (b) \(p_1\) and \(p_2\) are how strong the magnetic poles are, and \(\mu_0\) shows how easily magnetism goes through empty space. (c) The unit for pole strength 'p' is found to be "ampere-meter" by looking at the magnetic force formula or magnetic moment definition.

🎯 Exam Tip: Understand the analogy between Coulomb's law for electrostatic forces and the force between magnetic poles. Be able to identify the physical quantities involved (\(p_1, p_2, \mu_0\)) and derive the unit for pole strength ('p') from either the force equation or the definition of magnetic moment.

Question 26. A single-pole doesn't exist. Why?

Answer:

A single magnetic pole (monopole) cannot exist in isolation. Every magnet, no matter how small or how many times it's divided, always has both a north and a south pole. This is because magnetism arises from moving charges (currents) or intrinsic properties of particles (like electron spin), which naturally create dipoles.

In simple words: You cannot have just one magnetic pole. Magnets always come with both a north and a south pole, even if you cut them into tiny pieces.

🎯 Exam Tip: This is a fundamental principle of magnetism: magnetic monopoles do not exist (unlike electric charges which can exist as isolated positive or negative). Always remember that magnetic field lines form closed loops, implying the existence of both poles.

Question 27. Why a freely suspended magnet always points in the north-south direction?

Answer:

A freely suspended magnet aligns itself in the north-south direction because the Earth acts as a giant magnet. The Earth's magnetic north pole is located near its geographical south pole, and its magnetic south pole is near its geographical north pole. Therefore, the north pole of the suspended magnet is attracted to the Earth's magnetic south pole (near geographic north), and its south pole is attracted to the Earth's magnetic north pole (near geographic south), causing it to align along the geomagnetic north-south direction.

In simple words: A free magnet points north-south because Earth is a huge magnet. The north end of your magnet is pulled by Earth's magnetic south pole (which is near Earth's geographic north pole), and vice versa.

🎯 Exam Tip: Reinforce the concept that Earth itself is a large magnet, and opposite poles attract. This explains the alignment of a compass needle with Earth's magnetic field, not its geographic poles directly.

Question 28. When a bar magnet is dropped from the top of a building, its magnetism decreases. Why?

Answer:

When a bar magnet is dropped and repeatedly hits the ground, the impact causes the regular arrangement of its molecular magnets (or magnetic domains) to be disturbed. This disorganization leads to a reduction in the overall alignment of the domains, thereby decreasing the magnet's net magnetic strength or magnetism.

In simple words: Dropping a magnet makes its tiny internal magnets get jumbled up. This disorganizes them, reducing the magnet's overall strength.

🎯 Exam Tip: Relate the loss of magnetism to the disturbance of magnetic domains or molecular magnets within the material. Physical shocks, heating, and strong opposing magnetic fields can all demagnetize a material by disrupting this internal alignment.

Question 29. Is there any difference between magnetic lines of force and electric lines of force? If your answer is yes, what is that difference?

Answer:

Yes, there are significant differences between magnetic lines of force and electric lines of force.

One key difference is that magnetic lines of force always form closed loops. They originate from the north pole of a magnet, go into the south pole externally, and then continue inside the magnet from the south pole to the north pole. In contrast, electric lines of force do not form closed loops; they originate from positive charges and terminate on negative charges or extend to infinity.

In simple words: Yes, magnetic lines are different from electric lines. Magnetic lines always form complete circles, going out from one end of a magnet and looping back in. Electric lines start from positive charges and end on negative charges, they don't loop back to their start.

🎯 Exam Tip: The most crucial difference to remember is that magnetic field lines form closed loops, whereas electric field lines do not. This fundamental distinction reflects the non-existence of magnetic monopoles compared to electric charges.

Question 30. Magnetic lines of force never intersect each other. What is the reason?

Answer:

Magnetic lines of force never intersect each other because if they were to intersect at a point, it would imply that at that specific point, there would be two different directions for the magnetic field. However, the magnetic field at any given point can only have one unique direction. Therefore, intersection of magnetic field lines is physically impossible.

In simple words: Magnetic lines never cross because if they did, it would mean the magnetic force has two directions at one spot, which is impossible. Magnetic force can only point in one way at any given place.

🎯 Exam Tip: Emphasize that the tangent to a magnetic field line at any point gives the direction of the magnetic field at that point. If lines intersected, it would mean multiple directions at one point, which is physically impossible for a vector quantity like magnetic field.

Question 31. A bar magnet of length 2l and pole strength m is placed in a uniform magnetic field B, inclined at angle 'theta' with the field.

(a) What is the force acting on each pole?

(b) Why the magnet gets a rotating effect?

Answer:

(a) The force acting on each pole of the bar magnet in the uniform magnetic field B is given by \(F = mB\), where 'm' is the pole strength and 'B' is the magnetic induction (magnetic field strength).

(b) The magnet experiences a rotating effect (torque) because two equal and opposite forces, \(F = mB\), act on its north and south poles. These forces are applied at different points along the magnet, forming a couple. This couple creates a turning effect, or torque, which tends to align the magnet with the external magnetic field.

In simple words: (a) Each end of the magnet feels a force equal to its pole strength multiplied by the magnetic field strength. (b) The magnet spins because these two forces act in opposite directions but at different spots, creating a twist. This twist tries to line up the magnet with the magnetic field.

🎯 Exam Tip: Understand that in a uniform magnetic field, the net force on a magnetic dipole is zero, but a net torque acts on it (unless it's aligned with the field). The torque arises from equal and opposite forces acting on the poles, forming a couple.

Question 32. Show how a current loop acts as a magnetic dipole. Arrive at an expression for the magnetic dipole moment.

Answer:

A current loop acts as a magnetic dipole because it produces a magnetic field similar to that of a bar magnet. If a current-carrying loop is viewed from one side such that the current flows clockwise, that side behaves as a south pole. If viewed from the other side, where the current flows counter-clockwise, it behaves as a north pole. This dual polarity is characteristic of a magnetic dipole.

To derive the expression for the magnetic dipole moment (M) of a current loop:

Consider a planar loop of wire carrying a current 'I'.
The magnetic dipole moment (M) of a current loop is defined as the product of the current (I) flowing through the loop and the area (A) enclosed by the loop.

So, \(M = IA\).

For a coil with 'n' number of turns, if each turn has an area 'A' and carries current 'I', the total magnetic moment is \(M = nIA\).

The S.I. unit of magnetic moment is Ampere-meter squared (\(A \cdot m^2\)) or Joule per Tesla (\(J \cdot T^{-1}\)).

In simple words: A wire loop with current flowing through it behaves like a tiny bar magnet, having a north and a south pole. The strength of this "magnet" is called its magnetic dipole moment, which is calculated by multiplying the current in the loop by the area it encloses. If there are many loops, you multiply by the number of loops too.

🎯 Exam Tip: Clearly state the analogy: a current loop is a magnetic dipole. Remember the formula \(M = nIA\) (for N turns) and its S.I. unit. Be able to explain the polarity of the loop based on the direction of current flow (e.g., clockwise for south pole).

Question 33. From the above collection, two bodies show almost identical field patterns.

(a) Which are they?

(b) Draw the lines of force produced by them.

(c) Identify the cause producing the field in each.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र विभिन्न वस्तुओं को दिखाता है जो चुंबकीय क्षेत्र उत्पन्न करती हैं: (a) एक आवेश (इलेक्ट्रिक फील्ड), (b) एक बार चुंबक, (c) एक धारावाही लूप, और (d) एक विद्युत क्षेत्र में आवेश। प्रश्न इन वस्तुओं में से उन दो को पहचानने के लिए है जिनके चुंबकीय क्षेत्र पैटर्न लगभग समान हैं।

(a) (b) (c) (d)

Answer:

(a) The two bodies that show almost identical magnetic field patterns are (b) a bar magnet and (c) a current-carrying solenoid (or current loop if simplified).

(b)
ℹ️ चित्र व्याख्या (Diagram Explanation): चित्र में एक बार चुंबक और एक सोलेनोइड दोनों के चुंबकीय क्षेत्र रेखाएँ दिखाई गई हैं। बार चुंबक के लिए, रेखाएँ उत्तरी ध्रुव से निकलकर बाहरी रूप से दक्षिणी ध्रुव में प्रवेश करती हैं और चुंबक के अंदर दक्षिणी से उत्तरी ध्रुव तक जाती हैं, जिससे बंद लूप बनते हैं। सोलेनोइड के लिए, क्षेत्र रेखाएँ भी उत्तरी ध्रुव की तरह एक छोर से निकलकर बाहरी रूप से दूसरे छोर पर दक्षिणी ध्रुव की तरह प्रवेश करती हैं, और सोलेनोइड के अंदर सीधी चलती हैं, जिससे एक समान क्षेत्र का पैटर्न बनता है जो बार चुंबक के समान होता है।

(c) The cause producing the magnetic field in each is:

(b) Bar magnet: The magnetic field is caused by the alignment of tiny atomic magnets (magnetic domains) within the material.
(c) Solenoid (or Current Loop): The magnetic field is caused by the circulation of electric current in the wire loops, which creates an effective magnetic dipole.

In simple words: (a) A bar magnet and a coil of wire carrying electricity create very similar magnetic fields. (b) Both have magnetic lines that come out of one end and loop back into the other, just like a bar magnet. (c) The bar magnet's field comes from its internal tiny magnets lining up, while the coil's field comes from the moving electricity in its wires.

🎯 Exam Tip: Understand the equivalence between a current-carrying solenoid (or loop) and a bar magnet in terms of their magnetic field patterns. Recognize that the ultimate source of magnetism in both cases is moving charges (electron spins and orbits in a bar magnet, and current flow in a solenoid).

Question 34. In what way does Gauss's theorem in magnetostatics differ from the one in electrostatics? What conclusion do you draw from this?

Answer:

Gauss's theorem in magnetostatics states that the total magnetic flux passing through any closed surface is always zero. Mathematically, it is expressed as \(\oint \vec{B} \cdot d\vec{A} = 0\).

In contrast, Gauss's theorem in electrostatics states that the total electric flux passing through any closed surface is proportional to the net electric charge enclosed within that surface. Mathematically, it is expressed as \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0}\).

The key difference is that the net flux is zero for magnetism but non-zero for electrostatics if charge is enclosed. From this, we conclude that isolated magnetic poles (magnetic monopoles) do not exist. Magnetic field lines always form closed loops, meaning they have no beginning or end, unlike electric field lines which start from positive charges and end on negative charges.

In simple words: Gauss's law for magnetism says that no magnetic field lines ever start or end, so the total amount of magnetic field going through any closed surface is always zero. But for electricity, Gauss's law says that electric field lines can start or end on electric charges, so the total electric field going through a surface is only zero if there's no charge inside. This difference means there are no isolated magnetic poles, unlike electric charges.

🎯 Exam Tip: The fundamental difference between Gauss's laws for electricity and magnetism is the crucial insight that magnetic monopoles do not exist. Always remember the mathematical forms of both laws and what they imply about the nature of electric charges and magnetic poles.

Question 35. What do you know about the following terms?

(a) Isogonic line and Agonic line

(b) Isoclinic line and Aclinic line

(c) Isodynamic line

Answer:

(a) An isogonic line is an imaginary line drawn on a map that connects places having the same magnetic declination (the angle between the magnetic meridian and the geographic meridian). An agonic line is a special type of isogonic line that connects places where the magnetic declination is zero, meaning the magnetic meridian and geographic meridian coincide.

(b) An isoclinic line is an imaginary line connecting places on a map that have the same angle of dip (or magnetic inclination). An aclinic line is a special type of isoclinic line that connects places where the angle of dip is zero. This line is essentially the magnetic equator.

(c) An isodynamic line is an imaginary line on a map that connects places having the same horizontal component of Earth's magnetic field intensity.

In simple words: (a) Isogonic lines connect places with the same magnetic compass error, while agonic lines connect places where there is no compass error. (b) Isoclinic lines connect places with the same dip angle of a magnetic needle, while the aclinic line connects places where the needle stays flat (zero dip). (c) Isodynamic lines connect places where Earth's sideways magnetic pull is the same strength.

🎯 Exam Tip: Be able to define each term precisely, understanding what specific magnetic element (declination, dip, horizontal intensity) each line connects. These lines are important for mapping Earth's magnetic field and its variations.

Question 36. Show that the magnetic moment of an atom is \(M = \frac{1}{2} e \omega r^2\), where 'e' is the charge of an electron, '\(\omega\)' is the angular speed of the electron, and 'r' is the radius of the electron orbit.

Answer:

Consider an electron with charge 'e' revolving around the nucleus of an atom in a circular orbit of radius 'r'. Let 'T' be its period of revolution.

The current (I) generated by this orbiting electron is \(I = \frac{\text{charge}}{\text{time period}} = \frac{e}{T}\).

The angular speed (\(\omega\)) of the electron is related to the period by \(T = \frac{2\pi}{\omega}\).

Substituting T into the current equation: \(I = \frac{e}{2\pi/\omega} = \frac{e\omega}{2\pi}\).

The magnetic moment (M) of a current loop is given by \(M = IA\), where A is the area of the loop. For a circular orbit, \(A = \pi r^2\).

Therefore, \(M = \left(\frac{e\omega}{2\pi}\right) (\pi r^2)\).

Simplifying the expression, we get: \(M = \frac{1}{2} e \omega r^2\).

In simple words: An electron moving in a circle around an atom creates a tiny electric current. This current makes a magnetic field, and its strength, called magnetic moment, can be calculated. We find that this magnetic moment is half the electron's charge multiplied by its spinning speed and the square of its orbit's radius.

🎯 Exam Tip: This derivation connects classical electromagnetism (current loops, magnetic moment) with atomic physics (electron orbits). Remember the definitions of current from orbiting charge and angular speed. Practice manipulating these equations to arrive at the final expression.

Question 37. According to Bohr's postulate, angular momentum of an electron is \(mvr = \frac{nh}{2\pi}\). Using this formula, show that the magnetic moment of the atom is \(M = n\mu_B\). Here what is \(\mu_B\) and what is its value?

Answer:

According to Bohr's postulate, the angular momentum (L) of an electron in a stable orbit is quantized: \(L = mvr = \frac{nh}{2\pi}\), where m is the electron's mass, v is its velocity, r is the orbit radius, n is the principal quantum number (1, 2, 3, ...), and h is Planck's constant.

From the previous question, the magnetic moment (M) of an orbiting electron is \(M = \frac{1}{2} e \omega r^2\).
We also know that \(v = \omega r\), so \(\omega = \frac{v}{r}\).

Substituting \(\omega\) into the magnetic moment equation: \(M = \frac{1}{2} e \left(\frac{v}{r}\right) r^2 = \frac{1}{2} evr\).

Now, we can express \(vr\) using Bohr's postulate: \(vr = \frac{nh}{2\pi m}\).

Substitute this into the magnetic moment equation:

\(M = \frac{1}{2} e \left(\frac{nh}{2\pi m}\right)\)

\(M = n \left(\frac{eh}{4\pi m}\right)\).

Thus, \(M = n\mu_B\).

Here, \(\mu_B\) is the Bohr magneton, which is the fundamental unit of atomic magnetic dipole moment. It represents the smallest possible magnetic moment for an electron.

The value of \(\mu_B\) is given by: \(\mu_B = \frac{eh}{4\pi m}\).

Substituting the values of e (\(1.602 \times 10^{-19} \text{ C}\)), h (\(6.626 \times 10^{-34} \text{ J s}\)), and m (\(9.109 \times 10^{-31} \text{ kg}\)):

\(\mu_B \approx 9.27 \times 10^{-24} \text{ Am}^2\) (or \(\text{J/T}\)).

In simple words: Using Bohr's idea that electrons have specific orbits and energies, we can show that an atom's magnetic strength is a multiple of a basic magnetic unit called the Bohr magneton. The Bohr magneton is the smallest unit of magnetism an electron can have, calculated from the electron's charge, Planck's constant, and the electron's mass.

🎯 Exam Tip: This derivation links Bohr's quantization condition for angular momentum to the magnetic moment, leading to the concept of the Bohr magneton. Clearly define \(\mu_B\) and remember its physical significance as a fundamental unit of atomic magnetic moment and its approximate value.

Question 38. A rod of a magnetic material moves with very small velocity v as shown in Figure below, through a uniform magnetic field. Draw how the magnetic lines of force take shape if the magnetic material is (i) Ferromagnetic (ii) Paramagnetic and (iii) Diamagnetic.

ℹ️ चित्र व्याख्या (Diagram Explanation): चित्र एक चुंबकीय क्षेत्र से गुजरने वाली एक छड़ को दिखाता है। हमें यह दर्शाना है कि चुंबकीय क्षेत्र रेखाएँ कैसी दिखेंगी जब छड़ (i) लौह-चुंबकीय, (ii) अनुचुंबकीय, और (iii) प्रतिचुंबकीय सामग्री से बनी हो। इसमें एक समान चुंबकीय क्षेत्र में एक चुंबकीय सामग्री की छड़ की गति को दर्शाया गया है।

(a)

Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): चित्र में तीन प्रकार की चुंबकीय सामग्रियों - प्रतिचुंबकीय, अनुचुंबकीय, और लौह-चुंबकीय - से गुजरने वाली चुंबकीय क्षेत्र रेखाओं के पैटर्न को दर्शाया गया है। प्रतिचुंबकीय सामग्री के लिए, क्षेत्र रेखाएँ सामग्री से दूर मुड़ती हैं, जो एक प्रतिकर्षण प्रभाव दिखाती हैं। अनुचुंबकीय सामग्री के लिए, रेखाएँ सामग्री के माध्यम से थोड़ी केंद्रित होती हैं, जो एक कमजोर आकर्षण प्रभाव दिखाती हैं। लौह-चुंबकीय सामग्री के लिए, रेखाएँ सामग्री के अंदर अत्यधिक केंद्रित होती हैं, जो एक मजबूत आकर्षण प्रभाव दिखाती हैं और सामग्री के माध्यम से क्षेत्र को महत्वपूर्ण रूप से बढ़ाती हैं।

In simple words: When a magnetic material moves through a magnetic field, the field lines bend differently depending on the material. For diamagnetic materials, the lines spread away from the material. For paramagnetic materials, the lines slightly gather inside. For ferromagnetic materials, the lines gather very strongly inside, making the field much stronger there.

🎯 Exam Tip: Be able to draw and explain the behavior of magnetic field lines for diamagnetic, paramagnetic, and ferromagnetic materials. The key is to show how each material either repels, slightly attracts, or strongly concentrates the field lines within itself.

Question 39. The following terms find importance in magnetism. Explain them

(a) Magnetic permeability

(b) Magnetic intensity (H)

(c) Intensity of magnetisation (I)

(d) Magnetic susceptibility (\(X_m\))

Answer:

(a) Magnetic Permeability (\(\mu\)): This is a measure of how easily a magnetic field can pass through a material, or how much a material can support the formation of a magnetic field within itself. It is defined as the ratio of magnetic induction (B) to magnetizing field strength (H): \(\mu = \frac{B}{H}\). It determines how a material reacts to an applied magnetic field.

(b) Magnetic Intensity (H): Also known as the magnetizing field or magnetic field strength, H represents the external magnetic field that magnetizes a material. It is the field produced by the external currents (like Amperian currents in a solenoid). For a solenoid with 'n' turns per unit length carrying current 'I', \(H = nI\). It is the field that exists within a material due to external sources.

(c) Intensity of Magnetization (M): This quantifies the degree to which a material becomes magnetized when placed in an external magnetic field. It is defined as the net magnetic moment (m) acquired by the material per unit volume (V): \(M = \frac{m}{V}\). It indicates the material's internal magnetic response.

(d) Magnetic Susceptibility (\(\chi_m\)): This dimensionless quantity describes how susceptible a material is to becoming magnetized in an external magnetic field. It is the ratio of the intensity of magnetization (M) induced in the material to the magnetizing field strength (H) that caused it: \(\chi_m = \frac{M}{H}\). A positive \(\chi_m\) indicates attraction (paramagnetic/ferromagnetic), while a negative \(\chi_m\) indicates repulsion (diamagnetic).

In simple words: (a) Magnetic permeability tells us how easily a material lets magnetic lines pass through it. (b) Magnetic intensity (H) is the strength of the magnetic field applied from outside. (c) Intensity of magnetization (M) shows how much a material itself becomes magnetized from the inside. (d) Magnetic susceptibility shows how easily a material can be magnetized by an external field.

🎯 Exam Tip: Master the definitions and formulas for these four key magnetic terms. Understand their physical significance and how they relate to each other. These concepts are fundamental for classifying and understanding the behavior of different magnetic materials.

Question 40. The figure shows a solenoid carrying a current of I - ampere. Its area is 'A' and the number of turns n.

ℹ️ चित्र व्याख्या (Diagram Explanation): चित्र एक सोलेनोइड दिखाता है जिसमें I एम्पीयर की धारा प्रवाहित हो रही है, जिसका अनुप्रस्थ-काट क्षेत्र 'A' है और प्रति इकाई लंबाई में घुमावों की संख्या 'n' है।

(a) What is the flux density of the solenoidal field inside?
(b) If magnetic material in the form of a rod of area 'a' (a << A) is inserted into the solenoid; what is the magnetizing field strength?
(c) Find the total flux density inside the specimen
(d) Derive expressions for susceptibility and relative permeability.
(e) Classify magnetic substances based on permeability, susceptibility, and total permeability. Give at least two examples for each type.

Answer:

(a) The flux density of the solenoidal field inside (in vacuum) is \(B_0 = \mu_0 nI\), where \(\mu_0\) is the permeability of free space, 'n' is the number of turns per unit length, and 'I' is the current.

(b) When a magnetic material rod is inserted into the solenoid, the magnetizing field strength (H) is created by the current in the solenoid. It is given by \(H = nI\), where 'n' is the number of turns per unit length and 'I' is the current. This field is largely independent of the material inside.

(c) The total flux density (B) inside the specimen (magnetic material) is the sum of the magnetic flux density in vacuum (\(B_0\)) and the magnetic flux density due to the magnetization of the material itself (\(B_m\)). So, \(B = B_0 + B_m = \mu_0 H + \mu_0 M = \mu_0 (H+M)\).

(d) Derivations for susceptibility and relative permeability:

We know that \(B = \mu_0(H+M)\).

Also, \(B = \mu H\), where \(\mu\) is the absolute permeability of the material.
So, \(\mu H = \mu_0(H+M)\).

Divide by H: \(\mu = \mu_0 \left(1 + \frac{M}{H}\right)\).

The magnetic susceptibility (\(\chi_m\)) is defined as \(\chi_m = \frac{M}{H}\).

So, \(\mu = \mu_0(1+\chi_m)\).

The relative permeability (\(\mu_r\)) is defined as \(\mu_r = \frac{\mu}{\mu_0}\).

Substituting \(\mu\), we get: \(\mu_r = \frac{\mu_0(1+\chi_m)}{\mu_0}\).

Therefore, \(\mu_r = 1+\chi_m\).

(e) Classification of magnetic substances:

(i) Diamagnetic substances:

- Susceptibility (\(\chi_m\)) is small and negative (e.g., \(-1 \le \chi_m < 0\)).
- Relative permeability (\(\mu_r\)) is slightly less than 1 (e.g., \(0 < \mu_r < 1\)).
- Absolute permeability (\(\mu\)) is less than \(\mu_0\) (\(\mu < \mu_0\)).
- Examples: Bismuth, lead, gold, mercury, silver, water, alcohol, air, hydrogen.

(ii) Paramagnetic substances:

- Susceptibility (\(\chi_m\)) is small and positive (e.g., \(0 < \chi_m < \varepsilon\), where \(\varepsilon\) is a small positive value).
- Relative permeability (\(\mu_r\)) is slightly greater than 1 (e.g., \(1 < \mu_r < 1+\varepsilon\)).
- Absolute permeability (\(\mu\)) is slightly greater than \(\mu_0\) (\(\mu > \mu_0\)).
- Examples: Aluminium, chromium, oxygen, platinum, manganese, sodium.

(iii) Ferromagnetic substances:

- Susceptibility (\(\chi_m\)) is large and positive (e.g., \(\chi_m \gg 1\)).
- Relative permeability (\(\mu_r\)) is much greater than 1 (e.g., \(\mu_r \gg 1\)).
- Absolute permeability (\(\mu\)) is much greater than \(\mu_0\) (\(\mu \gg \mu_0\)).
- Examples: Iron, cobalt, nickel.

In simple words: (a) The magnetic field strength inside a solenoid (in empty space) is found by multiplying a constant (\(\mu_0\)), the turns per length, and the current. (b) If a magnetic rod is put inside, the force that magnetizes it (magnetizing field strength) is still just from the coil's current. (c) The total magnetic field inside the rod is the sum of the coil's field and the field made by the rod itself. (d) We found formulas that connect how easily a material gets magnetized (susceptibility) and how well it lets magnetic fields pass through (relative permeability) to its basic magnetic properties. (e) Materials are sorted into three groups: diamagnetic (weakly pushed away), paramagnetic (weakly pulled in), and ferromagnetic (strongly pulled in), based on these properties.

🎯 Exam Tip: This question covers several core concepts. For solenoids, remember \(B_0 = \mu_0 nI\) and \(H = nI\). Be able to derive the relation \(\mu_r = 1+\chi_m\). Crucially, know the characteristic ranges of \(\chi_m\), \(\mu_r\), and \(\mu\) for diamagnetic, paramagnetic, and ferromagnetic materials, along with examples for each.

Question 41. The relation between \(\mu_r\) and \(\chi_m\) is \(\mu_r = 1 + \chi_m\). How will you arrive at this relation? Explain.

Answer:

To derive the relation \(\mu_r = 1 + \chi_m\), we start with the definition of the total magnetic induction (B) inside a material when an external magnetizing field (H) is applied.

The total magnetic induction B in a material is the sum of the magnetic induction in free space (\(B_0\)) due to the external field H, and the magnetic induction (\(B_m\)) due to the magnetization M of the material itself.

So, \(B = B_0 + B_m\).

We know that \(B_0 = \mu_0 H\) (where \(\mu_0\) is the permeability of free space) and \(B_m = \mu_0 M\) (where M is the intensity of magnetization).

Substituting these into the equation for B: \(B = \mu_0 H + \mu_0 M = \mu_0 (H + M)\) ...(1)

Also, by definition, the total magnetic induction B in a material can be written as \(B = \mu H\) (where \(\mu\) is the absolute permeability of the material). ...(2)

Equating (1) and (2): \(\mu H = \mu_0 (H + M)\).

Divide both sides by H: \(\mu = \mu_0 \left(1 + \frac{M}{H}\right)\).

By definition, magnetic susceptibility (\(\chi_m\)) is \(\chi_m = \frac{M}{H}\).

Substituting \(\chi_m\) into the equation: \(\mu = \mu_0 (1 + \chi_m)\).

Finally, the relative permeability (\(\mu_r\)) is defined as \(\mu_r = \frac{\mu}{\mu_0}\).

So, \(\frac{\mu}{\mu_0} = 1 + \chi_m\).

Therefore, \(\mu_r = 1 + \chi_m\).

In simple words: We get the relationship \(\mu_r = 1 + \chi_m\) by understanding that the total magnetic field inside a material is made up of the external field and the field the material creates itself. By using definitions of total magnetic field, permeability of free space, and how much a material gets magnetized, we can mathematically connect relative permeability (how easily magnetism passes through it) and magnetic susceptibility (how easily it gets magnetized).

🎯 Exam Tip: This derivation is fundamental. Start with the total magnetic induction \(B = \mu_0(H+M)\) and the definition \(B = \mu H\). Clearly define \(\chi_m = M/H\) and \(\mu_r = \mu/\mu_0\). Ensure each step of substitution and algebraic manipulation is correct.

Question 42. On the basis of magnetic properties, the substances are classified into three.

(a) What are these three classifications?

(b) Explain each of them.

(c) Give three examples for each.

Answer:

(a) The three classifications of substances based on their magnetic properties are:

(i) Diamagnetic substances
(ii) Paramagnetic substances
(iii) Ferromagnetic substances

(b) Explanation of each classification:

(i) **Diamagnetic substances:** These materials do not possess a net magnetic moment in their atoms or molecules in the absence of an external magnetic field. When placed in an external magnetic field, they develop a weak magnetic moment in the direction opposite to the applied field. Consequently, they are weakly repelled by magnets and tend to move from stronger to weaker parts of a non-uniform magnetic field.

(ii) **Paramagnetic substances:** The individual atoms, molecules, or ions of these materials have a net non-zero magnetic moment even without an external magnetic field. However, these moments are randomly oriented, so the net magnetization is zero. When placed in an external magnetic field, the atomic magnetic moments partially align with the field, leading to a weak magnetization in the direction of the field. They are weakly attracted by magnets and tend to move from weaker to stronger parts of a non-uniform magnetic field.

(iii) **Ferromagnetic substances:** These materials have a strong net magnetic moment in their atoms/molecules. They are characterized by the presence of magnetic domains, which are regions where atomic magnetic moments are strongly aligned. When an external field is applied, these domains reorient and grow, leading to a very strong magnetization in the direction of the field. They are strongly attracted by magnets and tend to move from weaker to stronger parts of a non-uniform magnetic field. They also exhibit hysteresis.

(c) Examples for each type:

(i) **Diamagnetic materials:** Bismuth, lead, gold, mercury, silver, water, copper, alcohol, air, hydrogen.

(ii) **Paramagnetic materials:** Aluminium, platinum, chromium, oxygen, manganese, sodium, copper chloride, crown glass, liquid oxygen, salt solutions of iron and nickel.

(iii) **Ferromagnetic materials:** Iron, nickel, cobalt, gadolinium, and their alloys.

In simple words: (a) Materials are grouped into diamagnetic, paramagnetic, and ferromagnetic based on how they react to magnets. (b) Diamagnetic materials are slightly pushed away by magnets because their internal magnetism opposes the external field. Paramagnetic materials are slightly pulled in because their tiny internal magnets weakly line up with the external field. Ferromagnetic materials are strongly pulled in because their internal magnetic regions (domains) line up very strongly. (c) Examples are: Bismuth (diamagnetic), Aluminum (paramagnetic), and Iron (ferromagnetic).

🎯 Exam Tip: This is a crucial concept. Be able to define, explain the atomic/molecular origin of magnetism, and describe the macroscopic behavior (attraction/repulsion, field line interaction) for each of the three types of magnetic materials. Memorize at least 2-3 examples for each classification.

 

Question 43. Is there any importance for the Curie point? Explain.

Answer: The Curie point is important because it is a specific temperature. When a ferromagnetic material gets hotter, its ability to be easily magnetized (susceptibility) goes down. At the Curie temperature, a ferromagnetic material completely loses its ferromagnetic properties and changes into a paramagnetic material. For example, iron has a Curie temperature of about 1000 K, cobalt is about 1400 K, and nickel is about 631 K.
In simple words: The Curie point is a special temperature. Above this temperature, strong magnets become weak magnets.

🎯 Exam Tip: Understanding the Curie point is crucial for explaining how temperature affects magnetic materials, especially ferromagnets.

 

Question 44. A (hypothetical) bar magnet (AB) is cut into two equal parts. One part is now kept over the other, so that pole C2 is above C₁ If M is the magnetic moment of the original magnet, what would be the magnetic moment of the combination so formed?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक बार मैग्नेट AB को दिखाता है। इसे बीच से दो बराबर भागों में काटा गया है, जिससे कटे हुए सिरों पर C1 और C2 ध्रुव बनते हैं। एक भाग को दूसरे के ऊपर रखा गया है, जिसमें C2 ठीक C1 के ऊपर है। यह एक संयुक्त संरचना बनाता है।
Answer: When a bar magnet is cut into two equal pieces and placed one on top of the other, the resulting magnetic moment of this new arrangement will be zero. This is because the magnetic fields from the two pieces will cancel each other out.
In simple words: If you cut a magnet and stack the pieces, their magnetic effects cancel out, so the total magnetic strength becomes zero.

🎯 Exam Tip: Remember that cutting a magnet creates new poles, and stacking them in opposition can lead to cancellation of magnetic moments.

 

Question 45. Define the S.I unit of the magnetic field. "A charge moving at right angles to a uniform magnetic, the field does not undergo a change in kinetic energy." Why?

Answer: The SI unit for magnetic field is the Tesla. One Tesla is defined when a one coulomb charge, moving at one meter per second perpendicular to the magnetic field, feels a force of one Newton. The magnetic force acting on a moving charged particle is always at a 90-degree angle to its movement. This means the magnetic force does no work on the particle, so the particle's kinetic energy does not change.
In simple words: Tesla is the unit of magnetic field. A magnetic field does not make a moving charge faster or slower because it pushes the charge sideways, not forwards or backwards.

🎯 Exam Tip: Key definitions like the Tesla and the reason for no kinetic energy change in a magnetic field are fundamental.

 

Question 46. How does the (i) pole strength and (ii) magnetic moment of each part of a bar magnet change if it is cut into two equal pieces along its length?

Answer: If a bar magnet is cut lengthwise into two equal parts, each new part will still be a magnet. However, both its pole strength (how strong its poles are) and its magnetic moment (its overall magnetic strength) will become smaller than the original magnet.
In simple words: Cutting a magnet along its length makes each new piece a weaker magnet, with less pole strength and a smaller magnetic moment.

🎯 Exam Tip: Understand that cutting a magnet, even along its length, always results in new magnets, each with its own north and south pole.

 

Question 47. (a) Where on the earth's surface is the value of the vertical component of the earth's magnetic field zero?
(b) The horizontal component of the earth's magnetic field at a given place is \(0.4 \times 10^{-4}\) T and the angle of dip is 30°. Calculate the value of (i) vertical component (ii) the total intensity of the earth's magnetic field.

Answer: (a) The vertical part of the Earth's magnetic field becomes zero at the magnetic equator.
(b)
(i) Given the horizontal component of Earth's magnetic field (\(H_E\)) is \(0.4 \times 10^{-4}\) T and the angle of dip (\(I\)) is 30 degrees. The vertical component (\(Z_E\)) can be calculated using the formula \(Z_E = H_E \tan I\).
\[Z_E = 0.4 \times 10^{-4} \times \tan 30^\circ\]
\[Z_E = 0.4 \times 10^{-4} \times 0.577\]
So, \(Z_E \approx 0.23 \times 10^{-4}\) T.
(ii) The total magnetic field intensity of Earth (\(B_E\)) can be found using the horizontal component (\(H_E\)) and the angle of dip (\(I\)). The formula is \(B_E = \frac{H_E}{\cos I}\).
\[B_E = \frac{0.4 \times 10^{-4}}{\cos 30^\circ}\]
\[B_E = \frac{0.4 \times 10^{-4}}{0.866}\]
So, \(B_E \approx 0.46 \times 10^{-4}\) T.
In simple words: (a) The Earth's magnetic field goes sideways (horizontal) at the equator, so its up-down (vertical) part is zero there.
(b) We use math with the given sideways field and the dip angle to find the up-down field and the total magnetic field strength.

🎯 Exam Tip: Remember the relationship between the horizontal component, vertical component, and total intensity of Earth's magnetic field using trigonometric functions (tan and cos) and the angle of dip.

 

Question 48. A uniform magnetic field gets modified as shown when two specimens X and Y are placed in it.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र समानांतर रेखाओं द्वारा दर्शाए गए एक समान चुंबकीय क्षेत्र को दिखाता है। जब नमूना X इस क्षेत्र में रखा जाता है, तो चुंबकीय क्षेत्र रेखाएँ फैल जाती हैं और इससे दूर धकेल दी जाती हैं। जब नमूना Y रखा जाता है, तो चुंबकीय क्षेत्र रेखाएँ इसमें केंद्रित हो जाती हैं और अंदर की ओर खींची जाती हैं।
(i) Identify the two specimens X and Y
(ii) State the reason for the behaviour of the field lines in X and Y.

Answer: (i) Specimen X is a diamagnetic material. Specimen Y is a paramagnetic material.
(ii) When a diamagnetic material like X is put into an external magnetic field, the magnetic field lines are pushed away from it, or expelled. This makes the magnetic field weaker inside the material. On the other hand, when a paramagnetic material like Y is placed in an external magnetic field, the magnetic field lines get pulled into it, becoming more concentrated inside the material, which makes the field stronger there.
In simple words: (i) X is a diamagnet (it pushes magnetic lines away). Y is a paramagnet (it pulls magnetic lines in).
(ii) Diamagnets push magnetic lines out, making the field inside weaker. Paramagnets pull magnetic lines in, making the field inside stronger.

🎯 Exam Tip: Be able to identify diamagnetic and paramagnetic materials based on how they affect magnetic field lines and explain the underlying reason for their behavior.

 

Question 49. How will a dia, para, and ferromagnetic material behave when kept in a nonuniform external magnetic field? Give one example of each of these materials.

Answer: When placed in a magnetic field that is not uniform (meaning its strength varies), a diamagnetic material will slowly move from the stronger areas of the field to the weaker areas. A paramagnetic material will slowly move from the weaker parts of the field to the stronger parts. A ferromagnetic material will move quickly and easily from the weaker parts of the field towards the stronger parts.
Examples of these materials are:
- Diamagnetic material: Bismuth
- Paramagnetic material: Aluminium
- Ferromagnetic material: Iron
In simple words: - Diamagnetic things move away from strong magnetic fields. (Example: Bismuth)
- Paramagnetic things move towards strong magnetic fields, but slowly. (Example: Aluminium)
- Ferromagnetic things move towards strong magnetic fields quickly. (Example: Iron)

🎯 Exam Tip: Distinguish the behavior of diamagnetic, paramagnetic, and ferromagnetic materials in non-uniform magnetic fields, and provide common examples for each.

 

Question 50. Why does a paramagnetic substance display greater magnetisation for the same magnetising field when cooled? How does a diamagnetic substance respond to similar temperature changes?

Answer: When a paramagnetic substance is cooled down, its tiny atomic magnets (dipoles) find it easier to line up with the external magnetic field. This alignment results in the material showing a stronger overall magnetism (greater magnetization) at lower temperatures for the same applied field. In contrast, the magnetism of a diamagnetic substance does not change with temperature; its magnetization is not affected by cooling or heating.
In simple words: When a paramagnetic material gets cold, its small magnets line up better with the outside magnetic field, making it more magnetic. A diamagnetic material's magnetism does not change with temperature.

🎯 Exam Tip: Remember that paramagnetic behavior is temperature-dependent (increases when cooled), while diamagnetic behavior is generally independent of temperature.