Maharashtra Board Class 12 Physics Chapter 10 Magnetic Fields due to Electric Current Exercise Solutions

Choose the correct option.

(i) A conductor has 3 segments; two straight and of length L each and a semicircular with radius R. It carries a current I. What is the magnetic field B at point P?\
(a) \( \frac{\mu_0 I}{4 \pi R} \)
(b) \( \frac{\mu_0 I}{4 \pi R^2} \)
(c) \( \frac{\mu_0 I}{4 R} \)
(d) \( \frac{\mu_0 I}{4 \pi} \)
Answer: (c) \( \frac{\mu_0 I}{4 R} \)
In simple words: The straight parts of the wire point directly at point P, so they don't create any magnetic field there. Only the half-circle part creates a magnetic field, which is exactly half of the field a full circle would make.

📝 Teacher's Note: Remind students that any segment of a wire pointing directly towards or away from a point contributes zero magnetic field at that point due to the \( \sin \theta \) term in the Biot-Savart law being zero.

🎯 Exam Tip: Identify parts of the conductor that contribute zero field first to simplify the problem immediately.

(ii) Figure a, b show two Amperian loops associated with the conductors carrying current I in the sense shown. The \( \oint \vec{B} \cdot \vec{dl} \) in the cases a and b will be, respectively,
(a) \( -\mu_0 I, 0 \)
(b) \( \mu_0 I, 0 \)
(c) \( 0, \mu_0 I \)
(d) \( 0, -\mu_0 I \)
Answer: (a) \( -\mu_0 I, 0 \)
In simple words: Ampere's Law says the sum of magnetic field around a loop depends on the net current passing through it. In the first case, the direction of the current makes it negative, and in the second case, the current doesn't pass through the loop at all, so it's zero.

📝 Teacher's Note: Use the right-hand thumb rule to determine the positive direction of current relative to the chosen direction of the Amperian loop.

🎯 Exam Tip: Carefully check if the current actually pierces the area enclosed by the Amperian loop; if it's outside, the integral is zero.

(iii)  A proton enters a perpendicular uniform magnetic field B at origin along the positive x axis with a velocity v as shown in the figure. Then it will follow the following path. [The magnetic field is directed into the paper].
(a) It will continue to move along positive x axis.
(b) It will move along a curved path, bending towards positive x axis.
(c) It will move along a curved path, bending towards negative y axis.
(d) It will move along a sinusoidal path along the positive x axis.
Answer: (c) It will move along a curved path, bending towards negative y axis.
In simple words: When a positive charge like a proton moves through a magnetic field, it feels a sideways push. Using the right-hand rule, this push makes it curve downwards into the negative y-direction.

📝 Teacher's Note: Use Fleming's Left Hand Rule or the Right Hand Palm Rule to demonstrate the direction of force on a moving charge. Make sure to emphasize that protons are positive.

🎯 Exam Tip: Draw the velocity and field vectors separately on your scratchpad to avoid confusion when applying the cross-product rule.

(iv)  A conducting thick copper rod of length 1 m carries a current of 15 A and is located on the Earth’s equator. There the magnetic flux lines of the Earth’s magnetic field are horizontal, with the field of \( 1.3 \times 10^{-4} \) T, south to north. The magnitude and direction of the force on the rod, when it is oriented so that current flows from west to east, are
(a) \( 14 \times 10^{-4} \) N, downward.
(b) \( 20 \times 10^{-4} \) N, downward.
(c) \( 14 \times 10^{-4} \) N, upward.
(d) \( 20 \times 10^{-4} \) N, upward.
Answer: (d) \( 20 \times 10^{-4} \) N, upward.
In simple words: We calculate the force using the formula \( F = BIL \). By multiplying the magnetic field, current, and length, we get the force. The direction is determined by the right-hand rule, pointing up.

📝 Teacher's Note: This is a great real-world application of the motor effect. Use a compass and a pencil (as the rod) to help students visualize East-West and North-South directions in the classroom.

🎯 Exam Tip: Rounding might be needed in calculations, but usually, these multiple-choice questions have distinct enough answers that the correct choice stands out.

(v)  A charged particle is in motion having initial velocity \( \vec{V} \) when it enter into a region of uniform magnetic field perpendicular to \( \vec{V} \). Because of the magnetic force the kinetic energy of the particle will
(a) remain uncharged.
(b) get reduced.
(c) increase.
(d) be reduced to zero.
Answer: (a) remain uncharged.
In simple words: The magnetic force always pushes a moving charge sideways, never forward or backward. Since it doesn't push in the direction of motion, it changes the direction of the particle but not its speed or energy.

📝 Teacher's Note: Emphasize that the work done by a magnetic force is always zero because the force is always perpendicular to the displacement.

🎯 Exam Tip: Remember: Magnetic fields can change velocity (direction) but never speed or kinetic energy.

Question 2. A piece of straight wire has mass 20 g and length 1m. It is to be levitated using a current of 1 A flowing through it and a perpendicular magnetic field B in a horizontal direction. What must be the magnetic of B?
Answer:
Data: m = 20 g = \( 2 \times 10^{-2} \) kg, l = 1 m, I = 1 A,
g = 9.8 m/s²
To balance the wire, the upward magnetic force must be equal in magnitude to the downward force due to gravity.
\( \therefore F_m = IlB = mg \)
Therefore, the magnitude of the magnetic field,
\( B = \frac{mg}{Il} = \frac{(2 \times 10^{-2})(9.8)}{(1)(1)} = 0.196 \) T
In simple words: To make the wire float, the magnetic field must push it up with exactly the same strength that gravity pulls it down. By balancing these two forces, we find the required magnetic field strength is 0.196 Tesla.

📝 Teacher's Note: This experiment is called "Magnetic Levitation." Ask students what would happen if the current direction was reversed (the wire would be pushed down instead of up).

🎯 Exam Tip: Always convert units to SI (grams to kilograms) before starting calculations to avoid magnitude errors.

Question 3. Calculate the value of magnetic field at a distance of 2 cm from a very long straight wire carrying a current of 5 A (Given: \( \mu_0 = 4\pi \times 10^{-7} \) Wb/Am).
Answer:
Data : I = 5A, a = 0.02 m, \( \frac{\mu_0}{4\pi} = 10^{-7} \) T∙m/A
The magnetic induction,
\( B = \frac{\mu_0 I}{2\pi a} = \frac{\mu_0}{4\pi} \frac{2I}{a} = 10^{-7} \times \frac{2(5)}{2 \times 10^{-2}} = 5 \times 10^{-5} \) T
In simple words: A wire with electricity creates a magnetic field around it. The further you move away, the weaker it gets. At 2cm away from this specific wire, the field is very small, just 0.00005 Tesla.

📝 Teacher's Note: Draw the concentric magnetic field lines around the wire to visualize how the field weakens with distance (\( B \propto 1/r \)).

🎯 Exam Tip: Using the simplified constant \( \frac{\mu_0}{4\pi} = 10^{-7} \) makes your calculations much faster and less prone to pi-related errors.

Question 4. An electron is moving with a speed of \( 3.2 \times 10^6 \) m/s in a magnetic field of \( 6.00 \times 10^{-4} \) T perpendicular to its path. What will be the radius of the path? What will be frequency and the kinetic energy in keV ? [Given: mass of electron = \( 9.1 \times 10^{-31} \) kg, charge e = \( 1.6 \times 10^{-19} \) C, 1 eV = \( 1.6 \times 10^{-19} \) J]
Answer:
Data : \( v = 3 \times 10^7 \) m/s, \( B = 6 \times 10^{-4} \) T,
\( m_e = 9 \times 10^{-31} \) kg, \( e = 1.6 \times 10^{-19} \) C,
1 eV = \( 1.6 \times 10^{-19} \) J
The radius of the circular path,
\( r = \frac{m_e v}{|e| B} = \frac{(9 \times 10^{-31})(3 \times 10^7)}{(1.6 \times 10^{-19})(6 \times 10^{-4})} = \frac{2.7}{9.6} = 0.2812 \) m
The frequency of revolution,
\( f = \frac{|e| B}{2\pi m_e} = \frac{(1.6 \times 10^{-19})(6 \times 10^{-4})}{2 \times 3.142 \times (9 \times 10^{-31})} = \frac{9.6}{18 \times 3.142} \times 10^8 = 16.97 \) MHz
Since the magnetic force does not change the kinetic energy of the charge,
\( KE = \frac{1}{2} m_e v^2 = \frac{1}{2}(9 \times 10^{-31})(3 \times 10^7)^2 = \frac{81}{2} \times 10^{-17} \) J

\( \implies KE = \frac{81}{2(1.6 \times 10^{-19})} \times 10^{-17} \) eV

\( \implies KE = \frac{8.1}{3.2} \times 10^3 = 2.531 \) keV
In simple words: When an electron flies into a magnetic field, it starts going in circles. We can calculate how big that circle is, how many times it spins per second, and how much energy it has based on its speed and the field strength.

📝 Teacher's Note: Note that in the solution provided, the data for velocity used is \( 3 \times 10^7 \) m/s instead of the \( 3.2 \times 10^6 \) m/s mentioned in the question. Always check if the question's data matches the solution steps.

🎯 Exam Tip: For kinetic energy in keV, remember to divide the energy in Joules by \( 1.6 \times 10^{-19} \) to get eV first, then divide by 1000 for keV.

Question 5. An alpha particle (the nucleus of helium atom) (with charge +2e) is accelerated and moves in a vacuum tube with kinetic energy = 10.00 MeV. On applying a transverse a uniform magnetic field of 1.851 T, it follows a circular trajectory of radius 24.60 cm. Obtain the mass of the alpha particle. [charge of electron = \( 1.62 \times 10^{-19} \) C]
Answer:
Data: 1 eV = \( 1.6 \times 10^{-19} \) J,
\( E = 10 \text{ MeV} = 10^7 \times 1.6 \times 10^{-19} \text{ J} = 1.6 \times 10^{-12} \text{ J} \)
B = 1.88 T, r = 0.242 m, e = \( 1.6 \times 10^{-19} \) C
Charge of an \( \alpha \)-particle,
\( q = 2e = 2(1.6 \times 10^{-19}) = 3.2 \times 10^{-19} \) C

\( r = \frac{m_\alpha v}{qB} \) and \( E = \frac{1}{2} m_\alpha v^2 \)

\( \implies r^2 = \frac{(m_\alpha v)^2}{q^2 B^2} \) and \( 2Em_\alpha = (m_\alpha v)^2 \)

\( \implies r^2 = \frac{2Em_\alpha}{q^2 B^2} \)

\( \implies m_\alpha = \frac{(qBr)^2}{2E} = \frac{[(3.2 \times 10^{-19})(1.88)(0.242)]^2}{2(1.6 \times 10^{-12})} \)

\( \implies m_\alpha = (3.2 \times 10^{-26})(1.88 \times 0.242)^2 \)

\( \implies m_\alpha = (3.2 \times 10^{-26})(0.455)^2 \)

\( \implies m_\alpha = (3.2 \times 10^{-26})(0.2070) = 6.624 \times 10^{-27} \) kg
This gives the mass of the \( \alpha \)-particle.
[Note : The value of r has been adjusted to match with the answer. The CODATA (Committee on Data for Science and Technology) accepted value of \( m_\alpha \) is approximately \( 6.6446 \times 10^{-27} \) kg.]
In simple words: By knowing how much energy the alpha particle has and how sharply it curves in a magnetic field, we can weigh it. This heavy helium nucleus weighs about \( 6.624 \times 10^{-27} \) kg.

📝 Teacher's Note: This is the principle behind a Mass Spectrometer, which scientists use to identify atoms and molecules by measuring their mass-to-charge ratio.

🎯 Exam Tip: Be careful with the squaring step in the formula \( m = \frac{q^2 B^2 r^2}{2E} \). Squaring the product inside parentheses is a common place where students make calculation errors.

Question 6. Two wires shown in the figure are connected in a series circuit and the same amount of current of 10 A passes through both, but in apposite directions. Separation between the two wires is 8 mm. The length AB is S = 22 cm. Obtain the direction and magnitude of the magnetic field due to current in wire 2 on the section AB of wire 1. Also obtain the magnitude and direction of the force on wire 1. [\( \mu_0 = 4\pi \times 10^{-7} \) T.m/A]
Answer:
Data: \( I_1 = I_2 = 10 \) A, \( s = 8 \) mm = \( 8 \times 10^{-3} \) m, \( l = 0.22 \) m
By right hand grip rule, the direction of the magnetic field \( \vec{B_2} \) due to the current in wire 2 at AB is into the page and its magnitude is
\( B_2 = \frac{\mu_0}{4\pi} \frac{2I}{s} = 10^{-7} \times \frac{2(10)}{8 \times 10^{-3}} = \frac{1}{4} \times 10^{-3} \) T
The current in segment AB is upwards. Then, by Fleming’s left hand rule, the force on it due to \( \vec{B_2} \) is to the left of the diagram, i.e., away from wire 1, or repulsive. The magnitude of the force is
\( F_{\text{on 1 by 2}} = I_1 l B_2 = (10)(0.22) \times \frac{1}{4} \times 10^{-3} \)
\( \implies F = 5.5 \times 10^{-4} \) N
In simple words: When two wires have currents going in opposite directions, they push each other away. The magnetic field from one wire pushes on the other wire with a force of 0.00055 Newtons.

📝 Teacher's Note: This is a classic demonstration of "Like currents attract, opposite currents repel." Ask students to predict the direction if both currents went upwards.

🎯 Exam Tip: Don't forget to specify BOTH magnitude and direction for vectors like Force and Magnetic Field to get full marks.

Question 7. A very long straight wire carries a current 5.2 A. What is the magnitude of the magnetic field at a distance 3.1 cm from the wire? [ \( \mu_0 = 4\pi \times 10^{-7} \) T∙m/A]
Answer:
Data : I = 5.2 A, a = 0.031 m, \( \frac{\mu_0}{4\pi} = 10^{-7} \) T∙m/A
The magnetic induction,
\( B = \frac{\mu_0 I}{2\pi a} = \frac{\mu_0}{4\pi} \frac{2I}{a} = 10^{-7} \times \frac{2(5.2)}{3.1 \times 10^{-2}} \)
\( \implies B = 3.35 \times 10^{-5} \) T
In simple words: We are calculating the strength of the magnetic field near a power wire. Just over 3cm away from a 5.2 Amp current, the magnetic field strength is about 0.0000335 Tesla.

📝 Teacher's Note: Remind students that the formula used assumes the wire is "infinitely long." In real life, it works well as long as the distance 'a' is much smaller than the wire's length.

🎯 Exam Tip: Watch out for centimeters to meters conversion. Always double-check your decimal point placement.

Question 8. Current of equal magnitude flows through two long parallel wires having separation of 1.35 cm. If the force per unit length on each of the wires in \( 4.76 \times 10^{-2} \) N, what must be I ?
Answer:
Data : \( I_1 = I_2 = I, s = 1.35 \times 10^{-2} \text{ m, } \frac{F}{l} = 4.76 \times 10^{-2} \text{ N} \)
\( \frac{F}{l} = \left(\frac{\mu_0}{4\pi}\right) \frac{2I_1 I_2}{s} = \left(\frac{\mu_0}{4\pi}\right) \frac{2I^2}{s} \)
\( \therefore I^2 = \frac{\frac{F}{l} \cdot s}{2(\mu_0/4\pi)} \)
\( \implies I^2 = \frac{(4.76 \times 10^{-2}) \times 1.35 \times 10^{-2}}{2 \times 10^{-7}} = 3.213 \times 10^3 \)
\( \therefore I = \sqrt{32.13 \times 10^2} = 56.68 \) A
In simple words: If two parallel wires are pushing or pulling each other with a certain force, we can work backward to find out how much electricity is flowing through them. In this case, it takes a strong current of 56.68 Amps.

📝 Teacher's Note: This formula is actually how the "Ampere" was historically defined. It's a fundamental concept linking electricity and mechanical force.

🎯 Exam Tip: When you get a large number for \( I^2 \), adjust it to a power of 10 that has an even exponent (like \( 10^2 \)) to make finding the square root easier.

Question 9. Magnetic field at a distance 2.4 cm from a long straight wire is 16 \( \mu \)T. What must be current through the wire?
Answer:
Data: a = \( 2.4 \times 10^{-2} \) m, B = \( 1.6 \times 10^{-5} \) T,
\( \frac{\mu_0}{4\pi} = 10^{-7} \) T∙m/A
\( B = \frac{\mu_0 I}{2\pi a} = \frac{\mu_0}{4\pi} \frac{2I}{a} \)
The current through the wire,
\( I = \frac{1}{\mu_0/4\pi} \frac{aB}{2} = \frac{1}{10^{-7}} \frac{(2.4 \times 10^{-2})(1.6 \times 10^{-5})}{2} = 1.92 \) A
In simple words: By measuring the magnetic field near a wire, we can tell how much current is in it. For a field of 16 micro-Teslas at this distance, there must be about 1.92 Amps of current.

📝 Teacher's Note: Explain the prefix 'micro' (\( \mu \)) which means one-millionth (\( 10^{-6} \)). Students often forget to convert these units before solving.

🎯 Exam Tip: Rearrange the formula to solve for the unknown variable (I) before plugging in the numbers. This prevents algebraic mistakes during the calculation phase.

Question 10. The magnetic field at the centre of a circular current carrying loop of radius 12.3 cm is \( 6.4 \times 10^{-6} \) T. What will be the magnetic moment of the loop?
Answer:
Data: R = 12.3cm = \( 12.3 \times 10^{-2} \) m,
B = \( 6.4 \times 10^{-6} \) T, \( \mu_0 = 4\pi \times 10^{-7} \) T∙m/A
\( B = \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{2R} \times \frac{2\pi R^2}{2\pi R^2} = \frac{\mu_0}{4\pi} \frac{2I(A)}{R^3} \) \( (\because A = \pi R^2) \)

\( \implies B = \frac{\mu_0}{4\pi} \frac{2\mu}{R^3} \)

where \( \mu = IA \) is the magnetic dipole moment of the coil.
\( \therefore \mu = \frac{BR^3}{2(\mu_0/4\pi)} = \frac{(6.4 \times 10^{-6})(0.123)^3}{2 \times 10^{-7}} \)
\( \implies \mu = 5.955 \times 10^{-2} \) J/T (or A∙m²)
In simple words: A loop of current acts like a tiny magnet. By measuring its field at the center, we find its "magnetic strength" or moment, which is about 0.0596 units in this case.

📝 Teacher's Note: Show students how the magnetic field of a loop is similar to that of a bar magnet. This makes the concept of "magnetic moment" much easier to grasp.

🎯 Exam Tip: Use the magnetic moment formula \( \mu = IA \) and be careful with the units — both \( J/T \) and \( A\cdot m^2 \) are correct and acceptable.

Question 11. A circular loop of radius 9.7 cm carries a current 2.3 A. Obtain the magnitude of the magnetic field (a) at the centre of the loop and (b) at a distance of 9.7 cm from the centre of the loop but on the axis.
Answer:
Data: R = z = 9.7 cm = \( 9.7 \times 10^{-2} \) m, I = 2.3A, N = 1
(a) At the centre of the coil :
The magnitude of the magnetic induction,
\( B = \frac{\mu_0 NI}{2R} \)
\( \implies B = \frac{(4\pi \times 10^{-7})(1)(2.3)}{2(9.7 \times 10^{-2})} = \frac{2 \times 3.142 \times 2.3}{9.7} \times 10^{-5} \)
\( \implies B = 1.49 \times 10^{-5} \) T

(b) On the axis, at a distance z = 9.7 cm from the coil :
\( B = \frac{\mu_0}{4\pi} \frac{2\pi I R^2}{(R^2 + z^2)^{3/2}} \)
\( (R^2 + z^2)^{3/2} = (2R^2)^{3/2} = 2\sqrt{2} R^3 \) \( (\because R = z) \)
\( \therefore B = \frac{\mu_0}{4\pi} \frac{2\pi I R^2}{2\sqrt{2} R^3} = \frac{\mu_0}{4\pi} \frac{\pi I}{\sqrt{2} R} \)
\( \implies B = (10^{-7}) \frac{3.142 \times 2.3}{1.414 \times 9.7 \times 10^{-2}} \)
\( \implies B = \frac{7.227}{13.72} \times 10^{-5} = 5.267 \times 10^{-6} \text{ T} = 5.267 \mu\text{T} \)
In simple words: The magnetic field is strongest right in the middle of a loop. As you move away from it along a straight line through the center, the field gets weaker. In this problem, it drops from 14.9 to about 5.3 units.

📝 Teacher's Note: Use the axial field formula carefully. When \( z=R \), the denominator simplifies to \( 2\sqrt{2}R^3 \), which makes the calculation much easier to manage.

🎯 Exam Tip: Label your answers as part (a) and (b) clearly. Partial credit is often given for correctly identifying the change in formula between center and axis.

Question 12. A circular coil of wire is made up of 100 turns, each of radius 8.0 cm. If a current of 0.40 A passes through it, what be the magnetic field at the centre of the coil?
Answer:
Data: N = 100, R = \( 8 \times 10^{-2} \) m, I = 0.4A,
\( \mu_0 = 4\pi \times 10^{-7} \) T∙m/A
\( B = \frac{\mu_0 NI}{2R} \)
\( \implies B = \frac{(4\pi \times 10^{-7})(100)(0.4)}{2(8 \times 10^{-2})} = 3.142 \times 10^{-4} \) T
In simple words: Every turn of the wire adds to the magnetic field. With 100 turns all working together, we get a field strength of about 0.0003 Tesla.

📝 Teacher's Note: Compare this to Question 11 to show how the number of turns 'N' acts as a multiplier for the magnetic field strength.

🎯 Exam Tip: Multiply by 'N' at the very end or include it in your main formula from the start. Forgetting the 'N' is the most common mistake in coil problems.

Question 13. For proton acceleration, a cyclotron is used in which a magnetic field of 1.4 Wb/m² is applied. Find the time period for reversing the electric field between the two Ds.
Answer:
Data: B = 1.4 Wb/m², m = \( 1.67 \times 10^{-27} \) kg,
q = \( 1.6 \times 10^{-19} \) C
\( T = \frac{2\pi m}{qB} \)
\( t = \frac{T}{2} = \frac{\pi m}{qB} = \frac{(3.142)(1.67 \times 10^{-27})}{(1.6 \times 10^{-19})(1.4)} \)
\( \implies t = 2.342 \times 10^{-8} \) s
This is the required time interval.
In simple words: A cyclotron uses magnetic fields to spin particles and electric fields to kick them faster. The electric field must flip its direction exactly every time the particle finishes half a circle. This happens extremely fast — in about 23 billionths of a second!

📝 Teacher's Note: Explain that the particle moves in a semicircle within each 'D' of the cyclotron. Therefore, we need the time for half a cycle (\( T/2 \)), not the full period.

🎯 Exam Tip: Read carefully: "time period for reversing the field" refers to the time for one semi-circular path (\( t = T/2 \)). Using the full period \( T \) will lead to an incorrect answer.

Question 14. A moving coil galvanometer has been fitted with a rectangular coil having 50 turns and dimensions 5 cm × 3 cm. The radial magnetic field in which the coil is suspended is of 0.05 Wb/m². The torsional constant of the spring is \( 1.5 \times 10^{-9} \) Nm/degree. Obtain the current required to be passed through the galvanometer so as to produce a deflection of 30°.
Answer:
Data : N = 50, C = \( 1.5 \times 10^{-9} \) Nm/degree,
A = lb = 5 cm × 3 cm = 15 cm² = \( 15 \times 10^{-4} \) m²,
B = 0.05 Wb/m², \( \theta = 30^\circ \)
NIAB = C\( \theta \)
\( \therefore \) The current through the coil, \( I = \frac{C\theta}{NAB} \)
\( \implies I = \frac{1.5 \times 10^{-9} \times 30}{50 \times 15 \times 10^{-4} \times 0.05} = \frac{3 \times 10^{-5}}{5 \times 0.5} \)
\( \implies I = 1.2 \times 10^{-5} \) A
In simple words: A galvanometer is a sensitive device that turns a needle when electricity flows. For this specific one, it only takes a tiny amount of current (12 micro-Amps) to turn the needle by 30 degrees.

📝 Teacher's Note: Remind students that "radial magnetic field" ensures that the angle between the coil's area vector and the field is always \( 90^\circ \), simplifying the torque formula to \( \tau = NIAB \).

🎯 Exam Tip: Check the units for the torsional constant. If it's in per degree, keep the deflection in degrees. If it's in per radian, convert the deflection to radians.

Question 15. A solenoid of length \( \pi \) m and 5 cm in diameter has winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its centre along the axis.
Answer:
Data: L = 3.142 m, N = 1000, I = 5A,
\( \mu_0 = 4\pi \times 10^{-7} \) T∙m/A
The magnetic induction,
\( B = \mu_0 nI = \mu_0 \left(\frac{N}{L}\right) I \)
\( \implies B = (4\pi \times 10^{-7}) \left(\frac{1000}{3.142}\right)(5) = \frac{20 \times 3.142 \times 10^{-4}}{3.142} \)
\( \implies B = 2 \times 10^{-3} \) T
In simple words: A solenoid is like a long tube of wire coils. Inside this tube, the magnetic field is strong and steady. For this solenoid, the field at the very center is 0.002 Tesla.

📝 Teacher's Note: Explain that 'n' is the number of turns per unit length (\( N/L \)). The diameter doesn't affect the field inside an ideal solenoid, so it is extra information.

🎯 Exam Tip: Notice how \( \pi \) in the length cancels out with the \( \pi \) in \( \mu_0 \). Recognizing these cancellations early can save time during an exam.

Question 16. A toroid of narrow radius of 10 cm has 1000 turns of wire. For a magnetic field of \( 5 \times 10^{-2} \) T along its axis, how much current is required to be passed through the wire?
Answer:
Data : Central radius, r = 10 cm = 0.1 m, N = 1000,
B = \( 5 \times 10^{-2} \) T, \( \mu_0 = 4\pi \times 10^{-7} \) T∙m/A
The magnetic induction,
\( B = \frac{\mu_0 NI}{2\pi r} = \frac{\mu_0}{4\pi} \frac{2NI}{r} \)
The current through the wire,
\( I = \frac{1}{\mu_0/4\pi} \frac{rB}{2N} = \frac{1}{10^{-7}} \frac{(0.1)(5 \times 10^{-2})}{2(1000)} = 25 \) A
2irr 4ir r
\[ \therefore 5 \times 10^{-2} = 10^{-7} \times \frac{2(1000)I}{0.1} \]
\( \therefore I = \frac{50}{2} = 25A \)
This is the required current.
In simple words: A toroid is a solenoid bent into a donut shape. To get a magnetic field of 0.05 Tesla inside this donut, we need to pass a current of 25 Amps through the wire.

📝 Teacher's Note: Explain that a toroid is basically an endless solenoid. The magnetic field is completely contained within its circular core.

🎯 Exam Tip: The circumference \( 2\pi r \) acts as the length 'L' in the solenoid formula \( B = \mu_0 (N/L) I \). Keeping this relationship in mind helps if you forget the specific toroid formula.

Question 17. In a cyclotron protons are to be accelerated. Radius of its D is 60 cm. and its oscillator frequency is 10 MHz. What will be the kinetic energy of the proton thus accelerated? (Proton mass = \( 1.67 \times 10^{-27} \) kg, e = \( 1.60 \times 10^{-19} \) C, 1eV = \( 1.6 \times 10^{-19} \) J)
Answer:
Data : \( R = 0.6 \) m, \( f = 10^7 \) Hz, \( m_p = 1.67 \times 10^{-27} \) kg, \( e = 1.6 \times 10^{-19} \) C, 1 eV = \( 1.6 \times 10^{-19} \) J
\( f = \frac{qB}{2\pi m_p} \) and \( KE = \frac{q^2 B^2 R^2}{2m_p} \)

\( \implies f^2 = \frac{q^2 B^2}{4\pi^2 m_p^2} \)

\( \implies \frac{q^2 B^2}{2m_p} = 2\pi^2 f^2 m_p \)

\( \implies KE = 2\pi^2 f^2 m_p R^2 \)
\( = 2 \times 9.872 \times (10^7)^2 (1.67 \times 10^{-27})(0.6)^2 \)
\( = 11.87 \times 10^{-13} \) J
\( = \frac{11.87 \times 10^{-13}}{1.6 \times 10^{-19}} \) J
\( = 7.419 \times 10^6 \) eV
\( = 7.419 \) MeV
In simple words: To find the energy of the proton, we use its speed and the size of the cyclotron. By calculating how fast it spins and the magnetic force acting on it, we find that the proton gains 7.419 Million Electron Volts of energy.

📝 Teacher's Note: When solving cyclotron problems, always check the units of frequency. Converting MHz to Hz is a common step where students make errors. Remind students that energy in Joules must be divided by \( 1.6 \times 10^{-19} \) to get Electron Volts.

🎯 Exam Tip: Memorize the relationship between cyclotron frequency and kinetic energy directly (\( KE \propto f^2 \)) to save time during calculations involving magnetic field substitutions.

Question 18. A wire loop of the form shown in the figure carries a current I. Obtain the magnitude and direction of the magnetic field at P. Given : \( \frac{\mu_0 I}{4\pi R} \sqrt{2} \)
Answer:
The wire loop is in the form of a circular arc AB of radius R and a straight conductor BCA. The arc AB subtends an angle of \( \Phi = 270^{\circ} = \frac{3\pi}{2} \) rad at the centre of the loop P. Since \( PA = PB = R \) and C is the midpoint of AB, \( AB = \sqrt{2}R \) and \( AC = CB = \frac{\sqrt{2}R}{2} = \frac{R}{\sqrt{2}} \). Therefore, \( a = PC = \frac{R}{\sqrt{2}} \).
The magnetic inductions at P due to the arc AB and the straight conductor BCA are respectively,
\( B_1 = \frac{\mu_0}{4\pi} \frac{I\Phi}{R} \) and \( B_2 = \frac{\mu_0}{4\pi} \frac{2I}{a} \)
Therefore, the net magnetic induction at P is
\( B_{net} = B_1 + B_2 = \frac{\mu_0}{4\pi} I \left[ \frac{\Phi}{R} + \frac{2}{R/\sqrt{2}} \right] \)
\( = \frac{\mu_0}{4\pi} \frac{I}{R} \left[ \Phi + 2\sqrt{2} \right] = \frac{\mu_0}{4\pi} \frac{I}{R} \left[ \frac{3\pi}{2} + 2\sqrt{2} \right] \)
This is the required expression.
In simple words: The magnetic field at the center is made of two parts: one from the curved circular wire and one from the straight wire. We calculate each field separately using physics formulas and add them together to get the total strength.

📝 Teacher's Note: Use the right-hand thumb rule to show that both fields (from the arc and the straight segment) point in the same direction (into the page), which is why we add their magnitudes.

🎯 Exam Tip: Clearly show the conversion of the angle from degrees to radians (\( 270^{\circ} \) to \( 3\pi/2 \)) as the formula for arc magnetic field requires radians.

Question 19. Two long parallel wires going into the plane of the paper are separated by a distance R, and carry a current I each in the same direction. Show that the magnitude of the magnetic field at a point P equidistant from the wires and subtending angle \(\theta\) from the plane containing the wires, is \( B = \frac{\mu_0}{\pi} \frac{I}{R} \sin 2\theta \) What is the direction of the magnetic field?
Answer:
In above figure, \( \vec{B}_1 \) and \( \vec{B}_2 \) are the magnetic fields in the plane of the page due to the currents in wires 1 and 2, respectively. Their directions are given by the right hand grip rule: \( \vec{B}_1 \) is perpendicular to AP and makes an angle \( \Phi \) with the horizontal. \( \vec{B}_2 \) is perpendicular to BP and also makes an angle \( \Phi \) with the horizontal.
\( AP = BP = a = \frac{R/2}{\cos \theta} \)
and \( B_1 = B_2 = \frac{\mu_0}{4\pi} \frac{2I}{a} = \frac{\mu_0}{4\pi} \frac{2I(2 \cos \theta)}{R} = \frac{\mu_0}{\pi} \frac{I}{R} \cos \theta \)
Since the vertical components cancel out, the magnitude of the net magnetic induction at P is
\( B_{net} = 2B_1 \cos \Phi = 2B_1 \cos(90^{\circ} - \theta) = 2B_1 \sin \theta \)
\( = 2(\frac{\mu_0}{\pi} \frac{I}{R} \cos \theta) \sin \theta = \frac{\mu_0}{\pi} \frac{I}{R} \sin 2\theta \)
as required. \( \vec{B}_{net} \) is in the plane parallel to that of the wires and to the right as shown in the figure.
In simple words: When two wires carry current, they both create magnetic fields around them. At a point between them, parts of these fields cancel each other out while others add up, creating a total field that points horizontally.

📝 Teacher's Note: This is a classic vector addition problem. Emphasize that magnetic field vectors are always perpendicular to the line joining the wire and the point. Drawing the geometry carefully is key.

🎯 Exam Tip: Use the identity \( 2 \sin \theta \cos \theta = \sin 2\theta \) to simplify the final expression; examiners look for this specific trigonometric form in the proof.

Question 20. Figure shows a section of a very long cylindrical wire of diameter a, carrying a current I. The current density which is in the direction of the central axis of the wire varies linearly with radial distance r from the axis according to the relation \( J = J_0 r/a \). Obtain the magnetic field B inside the wire at a distance r from its centre. [Answer: \( B = \frac{\mu_0 J_0 r^2}{3a} \)]
Answer:
Consider an annular differential element of radius r and width dr. The current through the area dA of this element is
\( dI = JdA = (J_0 \frac{r}{a})2\pi r dr = \frac{2\pi J_0 r^2 dr}{a} \) ............ (1)
To apply the Ampere’s circuital law to the circular path of integration, we note that the wire has perfect cylindrical symmetry with all the charges moving parallel to the wire. So, the magnetic field must be tangent to circles that are concentric with the wire. The enclosed current is the current within radius r. Thus,
\( \oint B \cdot dl = \mu_0 I_{encl} \)

\( \implies \oint B dl = \mu_0 \int_0^r dI = \mu_0 \int_0^r \frac{2\pi J_0}{a} r^2 dr \) ............ (2)

\( \implies B(2\pi r) = \frac{\mu_0 2\pi J_0}{a} \left( \frac{r^3}{3} \right) \)

\( \implies B = \frac{\mu_0 J_0}{3a} r^2 \) ............ (3)
which is the required expression.
[Note : Integrating Eq. (1) from \( r=0 \) to \( r=a \),
\( I = \int dI = \int_0^a \frac{2\pi J_0}{a} r^2 dr = \frac{2\pi J_0}{a} \left[ \frac{r^3}{3} \right]_0^a = \frac{2\pi J_0 a^2}{3} \). Therefore,
\( J_0 = \frac{3I}{2\pi a^2} \), so that in terms of the total current \( I \) carried by the wire, \( B = \frac{\mu_0 r^2}{3 a} \left( \frac{3 I}{2\pi a^2} \right) = \frac{\mu_0 I}{2\pi a^3} r^2 \).]
In simple words: Inside a wire where electricity isn't flowing evenly, the magnetic field doesn't change linearly. By adding up all the small bits of current from the center to a certain distance, we find that the field increases with the square of that distance.

📝 Teacher's Note: Explain that since current density J depends on radius r, we cannot simply use \( I_{encl} = J \times \text{Area} \). Integration is necessary because the current is "denser" further away from the center.

🎯 Exam Tip: Always state the limits of integration clearly (\( 0 \) to \( r \)). If the question asks for the field in terms of total current \( I \), remember to perform the second integration from \( 0 \) to \( a \).

Question 21. In the above problem, what will be the magnetic field B inside the wire at a distance r from its axis, if the current density J is uniform across the cross section of the wire?
Answer:
Figure shows the cross section of a long straight wire of radius a that carries a current I out of the page. Because the current is uniformly distributed over the cross section of the wire, the magnetic field \( \vec{B} \) due to the current must be cylindrically symmetrical. Thus, along the Amperian loop of radius \( r(r < a) \), symmetry suggests that \( \vec{B} \) is tangent to the loop, as shown in the figure.
\( \oint \vec{B} \cdot \vec{dl} = B \oint dl = B(2\pi r) \) ........... (1)
Because the current is uniformly distributed, the current \( I_{encl} \) enclosed by the loop is proportional to the area encircled by the loop; that is,
\( I_{encl} = J\pi r^2 \)
By right-hand rule, the sign of \( dl \) is positive. Then, by Ampere’s law,
\( B (2\pi r) = \mu_0 I_{encl} = \mu_0 J\pi r^2 \) .................... (2)

\( \implies B = \frac{\mu_0 Jr}{2} \) ................. (3)
OR
\( I_{encl} = I \frac{\pi r^2}{\pi a^2} \)
By right-hand rule, the sign of \( I \frac{\pi r^2}{\pi a^2} \) is positive. Then, by Ampere’s law,
\( \oint B dl = \mu_0 I_{encl} \)

\( \implies B(2\pi r) = \mu_0 I \frac{r^2}{a^2} \) ................. (4)

\( \implies B = \left( \frac{\mu_0 I}{2\pi a^2} \right)r \) ................. (5)
[Note: Thus, inside the wire, the magnitude B of the magnetic field is proportional to distance r from the centre. At a distance r outside a straight wire, \( B = \frac{\mu_0 I}{2\pi r} \) i.e., \( B \propto \frac{1}{r} \).]
In simple words: If current flows evenly through a thick wire, the magnetic field gets stronger as you move from the center towards the surface. Once you go outside the wire, the field starts getting weaker as you move further away.

📝 Teacher's Note: Compare this result with the previous question. In uniform density, \( B \propto r \), but in variable density \( J \propto r \), we found \( B \propto r^2 \). This helps students see how the distribution of current affects the field shape.

🎯 Exam Tip: Draw the graph of B vs r for a thick wire (linear increase inside, hyperbolic decrease outside) to accompany this derivation; it demonstrates a deeper understanding to the examiner.

Theory Exercise

Question 1. Distinguish between the forces experienced by a moving charge in a uniform electric field and in a uniform magnetic field.
Answer:
A charge q moving with a velocity \( \vec{v} \) through a magnetic field of induction \( \vec{B} \) experiences a magnetic force perpendicular both to \( \vec{B} \) and \( \vec{v} \). Experimental observations show that the magnitude of the force is proportional to the magnitude of \( \vec{B} \), the speed of the particle, the charge q and the sine of the angle \( \theta \) between \( \vec{v} \) and \( \vec{B} \). That is, the magnetic force, \( F_m = qv B \sin \theta \)
\( \implies \vec{F}_m = q(\vec{v} \times \vec{B}) \)
Therefore, at every instant \( \vec{F}_m \) acts in a direction perpendicular to the plane of \( \vec{v} \) and \( \vec{B} \). If the moving charge is negative, the direction of the force \( \vec{F}_m \) acting on it is opposite to that given by the right-handed screw rule for the cross-product \( \vec{v} \times \vec{B} \).
If the charged particle moves through a region of space where both electric and magnetic fields are present, both fields exert forces on the particle.
The force due to the electric field \( \vec{E} \) is \( \vec{F}_e = q\vec{E} \).
The total force on a moving charge in electric and magnetic fields is called the Lorentz force :
\( \vec{F} = \vec{F}_e + \vec{F}_m = q(\vec{E} + \vec{v} \times \vec{B}) \)
Special cases :
(i) \( \vec{v} \) is parallel or antiparallel to \( \vec{B} \) : In this case, \( F_m = qvB \sin 0^{\circ} = 0 \). That is, the magnetic force on the charge is zero.
(ii) The charge is stationary \( (v = 0) \) : In this case, even if \( q \neq 0 \) and \( B \neq 0 \), \( F_m = q(0)B \sin \theta = 0 \). That is, the magnetic force on a stationary charge is zero.
In simple words: An electric field pulls a charge in the same direction as the field, even if the charge is standing still. A magnetic field only pushes a charge if it's moving, and it always pushes sideways (perpendicularly) to the motion.

📝 Teacher's Note: Use the analogy of a "wind" (electric field) vs a "banked curve on a road" (magnetic field). The electric field pushes you along, while the magnetic field only changes your direction if you are already moving.

🎯 Exam Tip: Highlight that the electric force is independent of velocity, whereas the magnetic force depends on both the magnitude and direction of velocity.

Question 2. Under what condition a charge undergoes uniform circular motion in a magnetic field? Describe, with a neat diagram, cyclotron as an application of this principle. Obtain an expression for the frequency of revolution in terms of the specific charge and magnetic field.
Answer:
Suppose a particle of mass m and charge q enters a region of uniform magnetic field of induction \( \vec{B} \). The magnetic force \( \vec{F}_m \) on the particle is always perpendicular to the velocity of the particle, \( \vec{v} \). Assuming the charged particle started moving in a plane perpendicular to \( \vec{B} \), its motion in the magnetic field is a uniform circular motion, with the magnetic force providing the centripetal acceleration.
If the charge moves in a circle of radius R,
\( F_m = |q|vB = \frac{mv^2}{R} \)

\( \implies mv = p = |q|BR \) ............. (1)
where p = mv is the linear momentum of the particle. Equation (1) is known as the cyclotron formula.
Frequency of revolution:
Since \( v = \omega R = (2\pi f) R \), substituting in \( qvB = \frac{mv^2}{R} \):
\( qB = \frac{mv}{R} = m(2\pi f) \)

\( \implies f = \frac{qB}{2\pi m} = \left( \frac{q}{m} \right) \frac{B}{2\pi} \)
where \( q/m \) is the specific charge.
In simple words: A charged particle moves in a perfect circle when it enters a magnetic field perpendicularly. This is used in machines like cyclotrons to spin particles faster and faster to study atoms.

📝 Teacher's Note: Students often forget the "perpendicular" condition. Emphasize that if the velocity is not perpendicular, the path becomes a helix (spiral), not a flat circle.

🎯 Exam Tip: When deriving frequency, show that it is independent of the radius \( R \) and velocity \( v \). This "isochronous" property is the most important feature of the cyclotron.

Question 3. What is special about a radial magnetic field ? Why is it useful in a moving coil galvanometer ?
Answer:
1. Advantage of radial magnetic field in a moving-coil galvanometer:

• As the coil rotates, its plane is always parallel to the field. That way, the deflecting torque is always a maximum depending only on the current in the coil, but not on the position of the coil.
• The restoring torque is proportional to the deflection so that a radial field makes the deflection proportional to the current. The instrument then has a linear scale, i.e., the divisions of the scale are evenly spaced. This makes it particularly straightforward to calibrate and to read.

2. Producing radial magnetic field :

• The pole pieces of the permanent magnet are made cylindrically concave, concentric with the axis of the coil.
• A soft iron cylinder is centred between the pole pieces so that it forms a narrow cylindrical gap in which the sides of the coil can move. Together, they produce a radial magnetic field; that is, the magnetic lines of force in the gap are along radii to the central axis.

In simple words: A radial field is special because it always points directly at the wire coil no matter how much the coil turns. This ensures the reading on the meter is perfectly proportional to the electricity flowing through it.

📝 Teacher's Note: Use a top-view diagram to show how the magnetic lines look like the spokes of a wheel. This visual helps students understand why the angle \( \theta \) between the area vector and the field stays \( 90^{\circ} \).

🎯 Exam Tip: Use the keyword "linear scale" when explaining the benefit. Examiners look for this to confirm you understand that current \( I \propto \theta \).

Question 4. State Biot-Savert law. Apply it to
(i) infinitely long current carrying conductor and
(ii) a point on the axis of a current carrying circular loop.
Answer: Consider a very short segment of length \( dl \) of a wire carrying a current \( I \). The product \( I d\vec{l} \) is called a current element; the direction of the vector \( d\vec{l} \) is along the wire in the direction of the current.
Biot-Savart law (Laplace law) : The magnitude of the incremental magnetic induction \( d\vec{B} \) produced by a current element \( I d\vec{l} \) at a distance \( r \) from it is directly proportional to the magnitude \( I \) of the current element, the sine of the angle between the current element \( Idl \) and the unit vector \( \hat{r} \) directed from the current element toward the point in question, and inversely proportional to the square of the distance of the point from the current element; the magnetic induction is directed perpendicular to both \( I d\vec{l} \) and \( \hat{r} \) as per the cross product rule.
\( dB \propto \frac{Idl \sin \theta}{r^2} \)
\( \therefore dB = \frac{\mu_0}{4\pi} \frac{Idl \sin \theta}{r^2} \) ........... (1)
In vector form, \( d\vec{B} = \left( \frac{\mu_0}{4\pi} \right) \frac{I d\vec{l} \times \hat{r}}{r^2} \) ........... (2)
where \( \hat{r} = \frac{\vec{r}}{r} \) and the constant \( \mu_0 \) is the permeability of free space. Equations (1) and (2) are called Biot-Savart law.
The incremental magnetic induction \( d\vec{B} \) is given by the right-handed screw rule of vector crossproduct \( I d\vec{l} \times \hat{r} \). In Fig, the current element \( I d\vec{l} \) and \( \hat{r} \) are in the plane of the page, so that \( d\vec{B} \) points out of the page at point P shown by \( \odot \); at the point Q, \( d\vec{B} \) points into the page shown by \( \otimes \).
The magnetic induction \( \vec{B} \) at the point due to the entire wire is, by the principle of superposition, the vector sum of the contributions \( d\vec{B} \) from all the current elements making up the wire.
From Eq. (2),
\( \vec{B} = \int d\vec{B} = \int \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2} \)
[Notes : (1) The above law is based on experiments by Jean Baptiste Biot (1774-1862) and Felix Savart (1791-1841), French physicists. From their observations Laplace deduced the law mathematically. (2) The Biot- Savart law plays a similar role in magnetostatics as Coulomb’s law does in electrostatics.]
In simple words: This law calculates the tiny bit of magnetic field created by a very small piece of wire carrying current. To find the total magnetic field of a long wire, we just add up all these tiny bits using math.

📝 Teacher's Note: Use the analogy of a small candle flame (current element) contributing to the total light in a room (magnetic field). Emphasize that \( d\vec{l} \) is a vector in the direction of conventional current.

🎯 Exam Tip: Always write the vector form of the law as it shows both magnitude and direction, which examiners specifically look for.

Question 5. State Ampere’s law. Explain how is it useful in different situations.
Answer: Ampere’s circuital law : In free space, the line integral of magnetic induction around a closed path in a magnetic field is equal to \( \mu_0 \) times the net steady current enclosed by the path.
In mathematical form,
\( \oint \vec{B} \cdot d\vec{l} = \mu_0 I \) ........... (1)
where \( \vec{B} \) is the magnetic induction at any point on the path in vacuum, \( d\vec{l} \) is the length element of the path, \( I \) is the net steady current enclosed and \( \mu_0 \) is the permeability of free space.
Explanation : Figure shows two wires carrying currents \( I_1 \) and \( I_2 \) in vacuum. The magnetic induction \( \vec{B} \) at any point is the net effect of these currents.
To find the magnitude \( B \) of the magnetic induction :
We construct an imaginary closed curve around the conductors, called an Amperian loop, and imagine it divided into small elements of length \( d\vec{l} \). The direction of \( dl \) is the direction along which the loop is traced.
(ii) We assign signs to the currents using the right hand rule : If the fingers of the right hand are curled in the direction in which the loop is traced, then a current in the direction of the outstretched thumb is taken to be positive while a current in the opposite direction is taken to be negative.
For each length element of the Amperian loop, \( \vec{B} \cdot d\vec{l} \) gives the product of the length \( dl \) of the element and the component of \( \vec{B} \) parallel to \( d\vec{l} \). If \( \theta \) is the angle between \( d\vec{l} \) and \( \vec{B} \),
\( \vec{B} \cdot d\vec{l} = (B \cos \theta) dl \)
Then, the line integral,
\( \oint \vec{B} \cdot d\vec{l} = \oint B \cos \theta dl \) ........... (2)
For the case shown in Fig., the net current \( I \) through the surface bounded by the loop is
\( I = I_2 - I_1 \)
\( \therefore \oint B \cos \theta dl = \mu_0 I \)
\( = \mu_0(I_2 - I_1) \) ........... (3)
Equation (3) can be solved only when \( B \) is uniform and hence can be taken out of the integral.
[Note : Ampere’s law in magnetostatics plays the part of Gauss’s law of electrostatics. In particular, for currents with appropriate symmetry, Ampere’s law in integral form offers an efficient way of calculating the magnetic field. Like Gauss’s law, Ampere’s law is always true (for steady currents), but it is useful only when the symmetry of the problem enables \( B \) to be taken out of the integral \( \oint \vec{B} \cdot d\vec{l} \). The current configurations that can be handled by Ampere’s law are infinite straight conductor, infinite plane, infinite solenoid and toroid.]
In simple words: Ampere's law relates the magnetic field around a loop to the total current passing through it. It's like a shortcut to calculate magnetic fields for simple shapes like long wires or coils.

📝 Teacher's Note: Compare this to walking around a track and counting how many people are crossing the field inside. It helps students understand the concept of 'enclosed' current.

🎯 Exam Tip: Remember that Ampere's Law is most effective when there is high symmetry (like a circle or rectangle) so the integration becomes simple multiplication.

Do you know (Textbook Page No. 230)

Question 1. You must have noticed high tension power transmission lines, the power lines on the big tall steel towers. Strong magnetic fields are created by these lines. Care has to be taken to reduce the exposure levels to less than 0.5 milligauss (mG).
Answer: With increasing population, many houses are constructed near high voltage overhead power transmission lines, if not right below them. Large transmission lines configurations with high voltage and current levels generate electric and magnetic fields and raises concerns about their effects on humans located at ground surfaces. With conductors typically 20 m above the ground, the electric field 2 m above the ground is about 0.2 kV / m to 1 kV / m. In comparison, that due to thunderstorms can reach 20 kV/m. For the same conductors, magnetic field 2 m above the ground is less than 6 \( \mu \)T. In comparison, that due to the Earth is about 40 \( \mu \)T.
In simple words: Big power lines create magnetic fields, but the field near the ground is actually much weaker than the Earth's natural magnetic field.

📝 Teacher's Note: Use this to address public health concerns. Explain that distance is the best shield against magnetic fields from power lines.

🎯 Exam Tip: Memorize the comparison values: 6 \( \mu \)T for power lines vs 40 \( \mu \)T for Earth to provide context in general questions.

Do you know (Textbook Page No. 232)

Question 1. Magnetic Resonance Imaging (MRI) technique used for medical imaging requires a magnetic field with a strength of 1.5 T and even upto 7 T. Nuclear Magnetic Resonance experiments require a magnetic field upto 14 T. Such high magnetic fields can be produced using superconducting coil electromagnet. On the other hand, Earth’s magnetic field on the surface of the Earth is about \( 3.6 \times 10^{-5} \) T = 0.36 gauss.
Answer: Magnetic Resonance Imaging (MRI) is a non-invasive imaging technology that produces three dimensional detailed anatomical images. Although MRI does not emit the ionizing radiation that is found in X-ray imaging, it does employ a strong magnetic field, e.g., medical MRIs usually have strengths between 1.5 T and 3 T.
The 21.1 T superconducting magnet at Maglab (Florida, US) is the world’s strongest MRI scanner used for Nuclear Magnetic Resonance (NMR) research. Since its inception in 2004, it has been continually conducting electric current of 284 A by itself. Because it is superconducting, the current runs through some 152 km of wire without resistance, so no outside energy source is needed. However, 2400 litres of liquid helium is cycled to keep the magnet at a superconducting temperature of 1.7 K. Even when not in use this magnet is kept cold; if it warms up to room temperature, it takes at least six weeks to cool it back down to operating temperature. The 45 T Hybrid Magnet of the Lab (which combines a superconducting magnet of 11.5 T with a resistive magnet of 33.5 T) is kept at 1.8 K using 2800 L of liquid helium and 15142 L of cold water.
In simple words: Doctors use MRI machines with huge magnets (much stronger than Earth's) to see inside the human body. These magnets are kept super cold so they can work without losing energy.

📝 Teacher's Note: Explain that "superconducting" means electricity flows perfectly without losing any energy as heat, which allows for such high magnetic fields.

🎯 Exam Tip: Note the unit conversion: 1 Tesla = 10,000 Gauss. This is a common objective question.

Question 2. Let us look at a charged particle which is moving in a circle with a constant speed. This is uniform circular motion that you have studied earlier. Thus, there must be a net force acting on the particle, directed towards the centre of the circle. As the speed is constant, the force also must be constant, always perpendicular to the velocity of the particle at any given instant of time. Such a force is provided by the uniform magnetic field \( \vec{B} \) perpendicular to the plane of the circle along which the charged particle moves.
Answer: When a charged particle moves in uniform circular motion inside a uniform magnetic field \( \vec{B} \) in a plane perpendicular to \( \vec{B} \), the centripetal force is the magnetic force on the particle. As in any UCM, this magnetic force is constant in magnitude and perpendicular to the velocity of the particle.
In simple words: When a charged particle enters a magnetic field, the field pushes it sideways. Because the push is always sideways to the direction it's moving, the particle ends up going in a perfect circle.

📝 Teacher's Note: Remind students of the "Right-hand rule" to determine the direction of force (\( \vec{F} = q\vec{v} \times \vec{B} \)).

🎯 Exam Tip: Mention that since the magnetic force is always perpendicular to velocity, the work done by the magnetic field on the particle is always zero.

Remember this (Textbook Page No. 233)

Question 1. Field penetrating into the paper is represented as \( \otimes \), while that coming out of the paper is shown by \( \odot \).
Answer: In a two-dimensional diagram, a vector pointing perpendicularly into the plane of the diagram is shown by a cross \( \otimes \) while that pointing out of the plane is shown by a dot \( \odot \) .
In simple words: Think of an arrow. If it's flying away from you (into the paper), you see the feathers (the cross). If it's coming at you (out of the paper), you see the sharp point (the dot).

📝 Teacher's Note: This visual aid is crucial for solving 3D physics problems on a flat sheet of paper. Draw an actual arrow on the board to illustrate this.

🎯 Exam Tip: Use these symbols in your diagrams to show magnetic field directions clearly; examiners expect these standard notations.

Do you know (Textbook Page No. 234)

Question 1. Particle accelerators are important for a variety of research purposes. Large accelerators are used in particle research. There have been several accelerators in India since 1953. The Department of Atomic Energy (DAE), Govt. of India, had taken initiative in setting up accelerators for research. Apart from ion accelerators, the DAE has developed and commissioned a 2 GeV electron accelerator which is a radiation source for research in science. This accelerator, ‘Synchrotron’, is fully functional at Raja Ramanna Centre for Advanced Technology, Indore. An electron accelerator, Microtron with electron energy 8-10 MeV is functioning at Physics Department, Savitribai Phule Pune University, Pune.
Answer: Particle accelerators are machines that accelerate charged subatomic particles to high energy for research and applications. They play a major role in the field of basic and applied sciences, in our understanding of nature and the universe. The size and cost of particle accelerators increase with the energy of the particles they produce. Medical Cyclotrons across the country are dedicated for medical isotope productions and for medical sciences. There are many existing and upcoming particle accelerators in India in different parts of country.
(https: / / www.researchgate.net/ publication/ 3209480 83_Existing and upcoming_particle_accelerators_in. India). For cutting-edge high energy particle physics, Indian particle physicists collaborate with those at Large Hadron Collider CERN, Geneva.
In simple words: These are giant machines that speed up tiny atoms to learn how the universe works and to create medicine for diseases like cancer.

📝 Teacher's Note: Explain that "acceleration" here means getting particles close to the speed of light using electric and magnetic fields.

🎯 Exam Tip: Be aware of the location of India's synchrotron accelerator (Indore) for general knowledge based science questions.

Can you recall (Textbook Page No. 238)

Question 1. How does the coil in a motor rotate by a full rotation? In a motor, we require continuous rotation of the current carrying coil. As the plane of the coil tends to become parallel to the magnetic field \( \vec{B} \), the current in the coil is reversed externally. Referring to Fig. the segment ab occupies the position cd. At this position of rotation, the current is reversed. Instead of from b to a, it flows from a to b, force \( \vec{F}_m \) continues to act in the same direction so that the torque continues to rotate the coil. The reversal of the current is achieved by using a commutator which connects the wires of the power supply to the coil via carbon brush contacts.
Answer: Electric Motor
From Fig. we see that the torque on a current loop rotates the loop to smaller values of \( \theta \) until the torque becomes zero, when the plane of the loop is perpendicular to the magnetic field and \( \theta = 0 \). If the current in the loop remains in the same direction when the loop turns past this position, the torque will reverse direction and turn the loop in the opposite direction, i.e., anticlockwise. To provide continuous rotation in the same sense, the current in the loop must periodically reverse direction, as shown in Fig.
In an electric motor, the current reversal is achieved externally by brushes and a split-ring commutator.
In simple words: In a motor, the wire loop keeps spinning because we flip the direction of electricity every half-turn. This flipping is done by a special part called a commutator.

📝 Teacher's Note: Demonstrate this with a simple hand-made motor if possible. Show how the loop would just wiggle back and forth without the commutator.

🎯 Exam Tip: The "commutator" is the keyword. In exams, focus on how it reverses current to maintain the same direction of torque.

Use your brain power (Textbook Page No. 242)

Question 1. Currents in two infinitely long, parallel wires exert forces on each other. Is this consistent with Newton’s third law?
Answer: Yes, they are equal in magnitude and opposite in direction and act on the contrary parts : \( \vec{F}_{\text{on 2 by 1}} = -\vec{F}_{\text{on 1 by 2}} \). Thus, they form action-reaction pair.
In simple words: If wire A pulls wire B with a certain force, wire B pulls wire A back with the exact same amount of force in the opposite direction.

📝 Teacher's Note: Reiterate that parallel currents attract and anti-parallel currents repel, but the forces between them are always an action-reaction pair.

🎯 Exam Tip: Use the formula \( F/L = \frac{\mu_0 I_1 I_2}{2\pi d} \) to prove that the magnitude of force per unit length is identical for both wires.

Do you know (Textbook Page No. 244)

Question 1. So far we have used the constant \( \mu_0 \) everywhere. This means in each such case, we have carried out the evaluation in free space (vacuum). \( \mu_0 \) is the permeability of free space.
Answer: Permeability of free space or vacuum, \( \mu_0 = 4\pi \times 10^{-7} \) H/m.
Earlier SI (2006) had fixed this value of \( \mu_0 \) as exact but revised SI fixes the value of e, requiring \( \mu_0 \) (and \( \varepsilon_0 \)) to be determined experimentally.
In simple words: This number tells us how easily a magnetic field can pass through empty space.

📝 Teacher's Note: Compare permeability (\( \mu \)) in magnetism to permittivity (\( \varepsilon \)) in electrostatics. Both represent how a medium responds to fields.

🎯 Exam Tip: Note the units of \( \mu_0 \): Henry per meter (H/m) or Tesla-meter per Ampere (T·m/A).

Use your brain power (Textbook Page No. 244)

Question 1. Using electrostatic analogue, obtain the magnetic field \( \vec{B}_{\text{equator}} \) at a distance d on the perpendicular bisector of a magnetic dipole of magnetic length 2l and moment \( \vec{M} \). For far field, verify that \( \vec{B}_{\text{equator}} = \left( \frac{\mu_0}{4\pi} \right) \frac{-\vec{M}}{(d^2+l^2)^{3/2}} \)
Answer: The magnitude of the electric intensity at a point at a distance r from an electric charge q in vacuum is given by
\( E = \frac{1}{4\pi\varepsilon_0} \frac{|q|}{r^2} \)
where \( \varepsilon_0 \) is the permittivity of free space. This intensity is directed away from the charge, if the charge is positive and towards the charge, if the charge is negative.
A magnetic pole is similar to an electric charge. The N-pole is similar to a positive charge and the S-pole is similar to a negative charge. Like an electric charge, a magnetic pole is assumed to produce a magnetic field in the surrounding region. The magnetic field at any point is denoted by a vector quantity called magnetic induction. Thus, by analogy, the magnitude of the magnetic induction at a point at a distance r from a magnetic pole of strength \( q_m \) is given by
\( B = \frac{\mu_0}{4\pi} \frac{q_m}{r^2} \)
This induction is directed away from the pole if it is an N-pole (strength \( +q_m \)) and towards the pole if it is an S-pole (strength \( -q_m \)).
Consider a point P on the equator of a magnetic dipole with pole strengths \( +q_m \) and \( -q_m \) and of magnetic length 2l. Let P be at a distance d from the centre of the dipole,
The magnetic induction at P due to the N-pole is directed along NP (away from the N-pole) while that due to the S-pole is along PS (towards the S-pole), each having a magnitude
\( B_N = B_S = \left( \frac{\mu_0}{4\pi} \right) \frac{q_m}{(d^2 + l^2)} \)
(\( \because NP = SP = \sqrt{d^2 + l^2} \))
The inductions due to the two poles are equal in magnitude so that the two, oppositely directed equatorial components, \( B_N \sin \theta \) and \( B_S \sin \theta \), cancel each other.
Therefore, the resultant induction is in a direction’ parallel to the axis of the magnetic dipole and has direction opposite to that of the magnetic moment of the magnetic dipole. The component of the induction due to the two poles along the axis is
\( 2 \left( \frac{\mu_0}{4\pi} \right) \frac{q_m}{(d^2+l^2)} \cos \theta \)
where \( \theta \) is the angle shown in the diagram. From the diagram,
\( \cos \theta = \frac{l}{(d^2+l^2)^{1/2}} \)
The total induction at P is then
\( B = (B_N + B_S) \cos \theta \)
\( = \left( \frac{\mu_0}{4\pi} \right) \frac{2q_m}{(d^2+l^2)} \cos \theta \)
\( = \left( \frac{\mu_0}{4\pi} \right) \frac{2q_m}{(d^2+l^2)} \frac{l}{(d^2+l^2)^{1/2}} \)
\( \therefore B = \left( \frac{\mu_0}{4\pi} \right) \frac{M}{(d^2 + l^2)^{3/2}} \)
where \( M = 2q_ml \) is the magnetic moment of the dipole.
\( \vec{B}_{\text{equator}} = \left( \frac{\mu_0}{4\pi} \right) \frac{-\vec{M}}{(d^2+l^2)^{3/2}} \)
The minus sign shows that the direction of \( \vec{B}_{\text{equator}} \) is opposite to that of \( \vec{M} \).
If the dipole is very short, i.e., \( l \ll d \), we can ignore \( l^2 \) in comparison with \( d^2 \). We then get
\( \vec{B}_{\text{equator}} = \left( \frac{\mu_0}{4\pi} \right) \frac{-\vec{M}}{d^3} \)
Thus, for a short dipole the induction varies inversely as the cube of the distance from it.
In simple words: To find the magnetic field of a bar magnet at its side, we treat the North and South poles like positive and negative charges. When you add their fields together at a side point, they partially cancel out, leaving a field that points backwards compared to the magnet's own direction.

📝 Teacher's Note: The "electrostatic analogue" is a powerful tool. North pole \( \approx +q \), South pole \( \approx -q \), and \( M \approx p \). Use this to help students remember formulas they already know from electricity.

🎯 Exam Tip: The \( 1/d^3 \) relationship is specifically for a "short dipole". In a general case, always use the full \( (d^2+l^2)^{3/2} \) denominator.

Question 2. What is the fundamental difference between an electric dipole and a magnetic dipole?
Answer: If a magnet is carefully and repeatedly cut, it would expose two new faces with opposite poles such that each piece would still be a magnet. This suggests that magnetic fields are essentially dipolar in character. The most elementary magnetic structure always behaves as a pair of two magnetic poles of opposite types and of equal strengths. Hence, analogous to an electric dipole, we hypothesize that there are positive and negative magnetic charges (or north and south poles) of equal strengths a finite distance apart within a magnet. Also, they are assumed to act as the source of the magnetic field in exactly the same way that electric charges act as the source of electric field. The magnitude of each ‘magnetic charge’ is referred to as its ‘pole strength’ and is equal to \( q_m = M / 2l \), where \( \vec{M} \) is the magnetic dipole moment, pointing from the negative (or south, S) pole to the positive (or north, N) pole.
However, while two types of electric charges exist in nature and have separate existence, isolated magnetic charges, or magnetic monopoles, are not observed. A magnetic pole is not an experimental fact: there are no real poles. To put it in another way, there are no point sources for \( \vec{B} \), as there are for \( \vec{E} \); there exists no magnetic analog to electric charge. Every experimental effort to demonstrate the existence of magnetic charges has failed. Hence, magnetic poles are called fictitious.
The electric field diverges away from a (positive) charge; the magnetic field line curls around a current. Electric field lines originate on positive charges and terminate on negative ones; magnetic field lines do not begin or end anywhere, they typically form closed loops or extend out to infinity.
In simple words: You can have a single positive or negative electric charge by itself. But you can never have just a North pole or just a South pole. If you cut a magnet in half, you just get two smaller magnets, each with its own North and South.

📝 Teacher's Note: Use the "cutting a magnet" demonstration. It’s the most intuitive way for students to grasp that magnetic monopoles don't exist in classical physics.

🎯 Exam Tip: The phrase "Magnetic monopoles do not exist" is a key technical point often required in descriptive answers about magnetism.

Do you know (Textbook Page No. 247)

Question 1. What is an ideal solenoid?
Answer: A solenoid is a long wire wound in the form of a helix. An ideal solenoid is tightly wound and infinitely long, i.e., its turns are closely spaced and the solenoid is very long compared to its crosssectional radius.
Each turn of a solenoid acts approximately as a circular loop. Suppose the solenoid carries a steady current I. The net magnetic field due to the current in the solenoid is the vector sum of the fields due to the current in all the turns. In the case of a tightly- wound solenoid of finite length, Fig. 10.39, the magnetic field lines are approximately parallel only near the centre of the solenoid, indicating a nearly uniform field there. However, close to the ends, the field lines diverge from one end and converge at the other end. This field distribution is similar to that of a bar magnet. Thus, one end of the solenoid behaves like the north pole of a magnet and the opposite end behaves like the south pole. The field outside is very weak near the midpoint.
For an ideal solenoid, the magnetic field inside is reasonably uniform over the cross section and parallel to the axis throughout the volume enclosed by the solenoid. The field outside is negligible in this case.
In simple words: An ideal solenoid is like a super-long slinky made of wire. When electricity runs through it, the magnetic field inside is perfectly straight and strong, while the field outside is almost zero.

📝 Teacher's Note: Compare a solenoid to a bar magnet. They create very similar field shapes, which helps students bridge the gap between electromagnetism and permanent magnets.

🎯 Exam Tip: Remember the property of an "ideal" solenoid: uniform field inside (\( B = \mu_0 n I \)) and zero field outside.

Use your brain power (Textbook Page No. 248)

Question 1. Choosing different Amperean loops, show that out-side an ideal toroid B = 0.
Answer: From below figure, the inner Amperean loop does not enclose any current while the outer Amperean loop encloses equal number of \( I_{in} \) and \( I_{out} \). Hence, by Ampere’s law, B = 0 outside an ideal toroid.
In simple words: If you draw a loop inside the hole of the doughnut (toroid) or outside the whole thing, the total current crossing that loop is zero. No net current means no magnetic field!

📝 Teacher's Note: Explain that for the outer loop, for every wire going "in" there is one coming "out", so they cancel each other out in the Ampere's Law sum.

🎯 Exam Tip: For any Amperean loop outside the toroid, clearly state that \( \sum I = 0 \), hence \( \oint \vec{B} \cdot d\vec{l} = 0 \).

Question 2. What is an ideal toroid?
Answer: A toroid is a toroidal solenoid. An ideal toroid consists of a long conducting wire wound tightly around a torus, a doughnut-shaped ring, made of a nonconducting material.
In an ideal toroid carrying a steady current, the magnetic field in the interior of the toroid is tangential to any circle concentric with the axis of the toroid and has the same value on this circle (the dashed line in figure). Also, the magnitude of the magnetic induction external to the toroid is negligible.
In simple words: Imagine taking a long solenoid and bending it until its ends meet to form a doughnut shape. That is a toroid. It traps the magnetic field entirely inside the doughnut.

📝 Teacher's Note: Toroids are used in things like tokamacs (fusion reactors) because they can contain magnetic fields without "leaky" ends.

🎯 Exam Tip: Note that in a toroid, the magnetic field lines are closed circles, unlike a straight solenoid where they are straight lines inside.