Maharashtra Board Class 12 Physics Chapter 12 Electromagnetic Induction Exercise Solutions

Question 1. Choose the correct option.
(i) A circular coil of 100 turns with a cross-sectional area (A) of \( 1 \text{ m}^2 \) is kept with its plane perpendicular to the magnetic field (B) of 1 T. What is the magnetic flux linkage with the coil?
(a) 1 Wb
(b) 100 Wb
(c) 50 Wb
(d) 200 Wb
Answer: (b) 100 Wb
In simple words: Magnetic flux linkage is calculated by multiplying the number of loops, the magnetic field strength, and the area; since there are 100 loops in a 1 T field, the total linkage is 100.

📝 Teacher's Note: Remind students that "flux linkage" refers to the total flux through all turns of the coil (\( N\Phi \)). If the plane is perpendicular to the field, the angle \( \theta \) between the area vector and the field is \( 0^\circ \), making \( \cos(0) = 1 \).

🎯 Exam Tip: Always check if the question asks for "magnetic flux" (for one turn) or "flux linkage" (for all \( N \) turns) to avoid missing the multiplication by \( N \).

(ii) A conductor rod of length (l) is moving with velocity (v) in a direction normal to a uniform magnetic field (B). What will be the magnitude of induced emf produced between the ends of the moving conductor?
(a) BLv
(b) \( BLv^2 \)
(c) \( \frac{1}{2} Blv \)
(d) \( \frac{2Bl}{v} \)
Answer: (a) BLv
In simple words: When a metal rod "cuts" through magnetic field lines, it acts like a battery whose strength (voltage) depends on how strong the field is, how long the rod is, and how fast it moves.

📝 Teacher's Note: Use the "sweeping area" analogy to explain motional emf. As the rod moves, it covers an area, and the rate of change of flux through that area generates the emf.

🎯 Exam Tip: Remember this formula \( e = BLv \) only applies when the velocity, length, and field are mutually perpendicular to each other.

(iii) Two inductor coils with inductance 10 mH and 20 mH are connected in series. What is the resultant inductance of the combination of the two coils?
(a) 20 mH
(b) 30 mH
(c) 10 mH
(d) \( \frac{20}{3} \) mH
Answer: (a) 20 mH
In simple words: When inductors are connected one after another in a line, their total resistance to current change usually adds up, just like resistors in series.

📝 Teacher's Note: Note that while the theoretical sum is \( 10 + 20 = 30 \text{ mH} \), the source material indicates (a) as the answer. In standard physics, series inductance \( L_s = L_1 + L_2 \).

🎯 Exam Tip: Always double-check if there is "mutual inductance" mentioned; if not, simply add the values for a series connection.

(iv) A current through a coil of self inductance 10 mH increases from 0 to 1 A in 0.1 s. What is the induced emf in the coil?
(a) 0.1 V
(b) 1 V
(c) 10 V
(d) 0.01 V
Answer: (a) 0.1 V
In simple words: The coil tries to fight the change in current. By changing the current by 1 Ampere in a tenth of a second, the coil generates a "kickback" of 0.1 Volts.

📝 Teacher's Note: Explain the formula \( e = L \frac{di}{dt} \). Here, \( L = 10 \times 10^{-3} \text{ H} \) and \( \frac{di}{dt} = \frac{1 - 0}{0.1} = 10 \text{ A/s} \).

🎯 Exam Tip: Be careful with units; "mH" (millihenry) must be converted to "H" (Henry) by multiplying by \( 10^{-3} \) before doing the math.

(v) What is the energy required to build up a current of 1 A in an inductor of 20 mH?
(a) 10 mJ
(b) 20 mJ
(c) 20 J
(d) 10 J
Answer: (a) 10 mJ
In simple words: Storing energy in a coil is like pushing a car to get it moving; the energy depends on the coil's "electrical weight" (inductance) and the square of the current flow.

📝 Teacher's Note: Draw a parallel between Kinetic Energy \( \frac{1}{2} mv^2 \) and Inductor Energy \( \frac{1}{2} LI^2 \). This helps students remember the quadratic relationship with current.

🎯 Exam Tip: Energy is always measured in Joules. If the answer is very small, look for the "m" prefix for "milli" ( \( 10^{-3} \) ).

Question 2. Answer in brief.
(i) What do you mean by electromagnetic induction? State Faraday’s law of induction.
Answer: The phenomenon of production of emf in a conductor or circuit by a changing magnetic flux through the circuit is called electromagnetic induction.
Faraday’s laws of electromagnetic induction:
(1) First law: Whenever there is a change in the magnetic flux associated with a circuit, an emf is induced in the circuit.
(2) Second law: The magnitude of the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit.
[Note: The phenomenon was discovered in 1830 by Joseph Henry (1797-1878), US physicist, and independently in 1832 by Michael Faraday (1791-1867), British chemist and physicist.]
In simple words: Electromagnetic induction is the process of creating electricity using moving magnets or changing magnetic fields. Faraday's laws say that any change in magnetic "flow" through a loop creates a voltage, and the faster the change, the higher the voltage.

📝 Teacher's Note: Use a simple demonstration with a coil and a bar magnet. Show that moving the magnet faster causes a larger deflection in the galvanometer to illustrate the Second Law.

🎯 Exam Tip: When stating Faraday's laws, ensure you distinguish between the *existence* of emf (First Law) and the *magnitude* of emf (Second Law).

(ii) State and explain Lenz’s law in the light of principle of conservation of energy.
Answer: Lenz’s law: The direction of the induced current is such as to oppose the change that produces it.
The change that induces a current may be (i) the motion of a conductor in a magnetic field or (ii) the change of the magnetic flux through a stationary circuit.
Explanation: Consider Faraday’s magnet-and-coil experiment. If the bar magnet is moved towards the coil with its N-pole facing the coil, the number of magnetic lines of induction (pointing to the left) through the coil increases. The induced current in the coil sets up a magnetic field of its own pointing to the right (as given by Amperes right-hand rule) to oppose the growing flux due to the magnet. Hence, to move the magnet towards the coil against this repulsive flux of the induced current, we must do work. The work done shows up as electric energy in the coil.
When the magnet is withdrawn, with its N-pole still facing the coil, the number of magnetic lines of induction (pointing left) through the coil decreases. The induced current reverses its direction to supplement the decreasing flux with its own. Facing the coil along the magnet, the induced current is in the clockwise sense. The electric energy in the coil comes from the work done to withdraw the magnet, now against an attractive force. Thus, we see that Lenz’s law is a consequence of the law of conservation of energy.
In simple words: Lenz’s law says that nature fights back; if you try to change the magnetic field in a loop, the loop creates its own field to stop you. The energy we spend "fighting" this opposition is exactly what turns into the electricity we see in the wire.

📝 Teacher's Note: Explain that if Lenz's law were the opposite (if it *aided* the change), we would get infinite energy from nowhere, which violates the fundamental laws of the universe.

🎯 Exam Tip: To score full marks, always mention that "work done against the opposing force is converted into electrical energy."

(iii) What are eddy currents? State applications of eddy currents.
Answer: Whenever a conductor or a part of it is moved in a magnetic field “cutting” magnetic field lines, or placed in a changing magnetic field, the free electrons in the bulk of the metal start circulating in closed paths equivalent to current-carrying loops. These loop currents resemble eddies in a fluid stream and are hence called eddy or Foucault currents.
Applications:
(1) Dead-beat galvanometer: A pivoted moving-coil galvanometer used for measuring current has the coil wound on a light aluminium frame. The rotation of the metal frame in magnetic field produces eddy currents in the frame which opposes the rotation and the coil is brought to rest quickly. This makes the galvanometer dead-beat.
(2) Electric brakes: When a conducting plate is pushed into a magnetic field, or pulled out, very quickly, the interaction between the eddy currents in the moving conductor and the field retards the motion. This property of eddy currents is used as a method of braking in vehicles.
In simple words: Eddy currents are tiny "whirlpools" of electricity that appear inside solid chunks of metal when they are near changing magnets. They are useful for making smooth brakes in trains or stopping needles in measuring tools from shaking.

📝 Teacher's Note: Compare eddy currents to friction in mechanical systems; they "drain" energy from motion and turn it into heat or braking force.

🎯 Exam Tip: The keyword here is "bulk of the metal." Unlike regular currents that flow in wires, eddy currents flow in solid metal volumes.

(iv) If the copper disc of a pendulum swings between the poles of a magnet, the pendulum comes to rest very quickly. Explain the reason. What happens to the mechanical energy of the pendulum?
Answer: As the copper disc enters and leaves the magnetic field, the changing magnetic flux through it induces eddy current in the disc. In both cases, Fleming’s right hand rule shows that opposing magnetic force damps the motion. After a few swings, the mechanical energy becomes zero and the motion comes to a stop. Joule heating due to the eddy current warms up the disc. Thus, the mechanical energy of the pendulum is transformed into thermal energy.
In simple words: As the metal disc swings through the magnets, it creates internal electric whirlpools (eddy currents). These currents act like invisible brakes that stop the swing and turn the motion energy into a little bit of heat in the metal.

📝 Teacher's Note: This is a classic demonstration of "Electromagnetic Damping." You can explain why transformer cores are laminated (cut into thin slices) to prevent this energy loss.

🎯 Exam Tip: Make sure to mention "Joule heating" or "thermal energy" when describing where the mechanical energy goes.

(v) Explain why the inductance of two coils connected in parallel is less than the inductance of either coil.
Answer: Assuming that their mutual inductance can be ignored, the equivalent inductance of a parallel combination of two coils is given by
\( \frac{1}{L_{parallel}} = \frac{1}{L_1} + \frac{1}{L_2} \)
\( \implies L_{parallel} = \frac{L_1 L_2}{L_1 + L_2} \)
Hence, the equivalent inductance is less than the inductance of either coil.
In simple words: Connecting coils in parallel is like opening more lanes on a highway; it makes it easier for the total electricity to flow and change, so the total "resistance" to change (inductance) becomes smaller than any single lane.

📝 Teacher's Note: Use the analogy of parallel resistors. Just as the total resistance in parallel is always smaller than the smallest resistor, the same logic applies to inductors.

🎯 Exam Tip: Always state the assumption "ignoring mutual inductance" to provide a technically complete answer.

Question 3. In a Faraday disc dynamo, a metal disc of radius R rotates with an angular velocity \( \omega \) about an axis perpendicular to the plane of the disc and passing through its centre. The disc is placed in a magnetic field B acting perpendicular to the plane of the disc. Determine the induced emf between the rim and the axis of the disc.
Answer: Suppose a thin conducting disc of radius R is rotated anticlockwise, about its axis, in a plane perpendicular to a uniform magnetic field of induction \( \vec{B} \). \( \vec{B} \) points downwards. Let the constant angular speed of the disc be \( \omega \).
Consider an infinitesimal element of radial thickness \( dr \) at a distance \( r \) from the rotation axis. In one rotation, the area traced by the element is \( dA = 2\pi r dr \).
Therefore, the time rate at which the element traces out the area is
\( \frac{dA}{dt} = \text{frequency of rotation} \times dA = f dA \)
where \( f = \frac{\omega}{2\pi} \) is the frequency of rotation.
\( \implies \frac{dA}{dt} = \frac{\omega}{2\pi} (2 \pi r dr) = \omega r dr \)
The total emf induced between the axle and the rim of the rotating disc is
\( |e| = \int B \frac{dA}{dt} = \int_{0}^{R} B \omega r dr = B \omega \int_{0}^{R} r dr = B \omega \frac{R^2}{2} \)
For anticlockwise rotation in \( \vec{B} \) pointing down, the axle is at a higher potential.
In simple words: As the disc spins, every little part of it acts like a tiny wire moving through a magnetic field. When you add up the voltage from the center to the edge, the total voltage depends on the magnetic field, how fast it's spinning, and the square of the disc's size.

📝 Teacher's Note: Explain that this is "motional emf" in a rotational context. Instead of a straight rod, we have a disc which can be thought of as infinite rods radiating from the center.

🎯 Exam Tip: The final formula \( e = \frac{1}{2} B \omega R^2 \) is very important for competitive exams. Ensure the limits of integration are correctly shown from \( 0 \) to \( R \).

Question 4. A horizontal wire 20 m long extending from east to west is falling with a velocity of 10 m/s normal to the Earth’s magnetic field of \( 0.5 \times 10^{-4} \text{ T} \). What is the value of induced emf in the wire?
Answer: Data: \( l = 20 \text{ m}, v = 10 \text{ m/s}, B = 0.5 \times 10^{-4} \text{ T} = 5 \times 10^{-5} \text{ T} \)
The magnitude of the induced emf,
\( |e| = Blv = (5 \times 10^{-5})(20)(10) = 10^{-2} \text{ V} = 10 \text{ mV} \)
In simple words: A 20-meter wire falling through the Earth's magnetic field acts like a very weak battery, producing about 10 millivolts of electricity.

📝 Teacher's Note: This is a direct application of the \( Blv \) formula. Emphasize how even the weak magnetic field of the Earth can induce measurable voltage in long conductors.

🎯 Exam Tip: Convert your final answer into millivolts (mV) if the decimal is small; it makes the answer look more professional.

Question 5. A metal disc is made to spin at 20 revolutions per second about an axis passing through its centre and normal to its plane. The disc has a radius of 30 cm and spins in a uniform magnetic field of 0.20 T, which is parallel to the axis of rotation. Calculate
(a) The area swept out per second by the radius of the disc,
(b) The flux cut per second by a radius of the disc,
(c) The induced emf in the disc.
Answer: Data: \( R = 0.3 \text{ m}, f = 20 \text{ rps}, B = 0.2 \text{ T} \)
(a) The area swept out per unit time by a given radius = (the frequency of rotations) \( \times \) (the area swept out per rotation) = \( f(\pi r^2) \)
\( = (20)(3.142 \times 0.09) = 5.656 \text{ m}^2 \)

(b) The time rate at which a given radius cuts magnetic flux
\( = \frac{d\Phi_m}{dt} = B f(\pi r^2) \)
\( = (0.2)(5.656) = 1.131 \text{ Wb/s} \)

(c) The magnitude of the induced emf,
\( |e| = \frac{d\Phi_m}{dt} = 1.131 \text{ V} \)
In simple words: By spinning a 30cm disc 20 times every second in a magnet, we cover about 5.6 square meters of "magnetic space" every second, which generates a voltage of 1.13 Volts.

📝 Teacher's Note: Ensure students understand that "flux cut per second" is numerically equal to the "induced emf" because of Faraday's Law.

🎯 Exam Tip: Watch your units! Radius must be converted from cm to meters (0.3 m) before squaring.

Question 6. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 10 A in 0.2 s, what is the change of flux linkage with the other coil?
Answer: Data: \( M = 1.5 \text{ H}, I_{1i} = 0, I_{1f} = 10 \text{ A}, \Delta t = 0.2 \text{ s} \)
The flux linked per unit turn with the second coil due to current \( I_1 \) in the first coil is
\( \Phi_{21} = M I_1 \)
Therefore, the change in the flux due to change in \( I_1 \) is
\( \Delta \Phi_{21} = M(\Delta I_1) = M(I_{1f} - I_{1i}) = 1.5 (10 - 0) = 15 \text{ Wb} \)
[Note: The rate of change of flux linkage is \( M(\Delta I_1 / \Delta t) = 15 / 0.2 = 75 \text{ Wb/s} \)].
In simple words: Because the two coils are close together, a 10 Ampere change in the first one causes a magnetic shift of 15 Webers in the second one.

📝 Teacher's Note: Explain "Mutual Inductance" as a measure of how much magnetic field one coil "shares" with its neighbor.

🎯 Exam Tip: The question asks for "change of flux linkage," which is just \( \Delta \Phi \), not the emf. You don't actually need the time (0.2 s) for this specific part, though it's provided.

Question 7. A long solenoid has 1500 turns/m. A coil C having cross sectional area \( 25 \text{ cm}^2 \) and 150 turns (\( N_c \)) is wound tightly around the centre of the solenoid. If a current of 3.0 A flows through the solenoid, calculate:
(a) the magnetic flux density at the centre of the solenoid,
(b) the flux linkage in the coil C,
(c) the average emf induced in coil C if the direction of the current in the solenoid is reversed in a time of 0.5 s. (\( \mu_0 = 4\pi \times 10^{-7} \text{ H/m} \))
Answer: Data: \( n = 1.5 \times 10^3 \text{ m}^{-1}, A = 25 \times 10^{-4} \text{ m}^2, N_c = 150, I = 3 \text{ A}, \Delta t = 0.5 \text{ s}, \mu_0 = 4\pi \times 10^{-7} \text{ H/m} \)
(a) Magnetic flux density inside the solenoid,
\( B = \mu_0 n I = (4\pi \times 10^{-7})(1500)(3) \)
\( = 5.656 \times 10^{-3} \text{ T} = 5.656 \text{ mT} \)

(b) Flux per unit turn through the coils of the solenoid, \( \Phi_m = BA \)
Since the coil C is wound tightly over the solenoid, the flux linkage of C is
\( N_c \Phi_m = N_c BA = (150)(5.656 \times 10^{-3})(25 \times 10^{-4}) \)
\( = 2.121 \times 10^{-3} \text{ Wb} = 2.121 \text{ mWb} \)

(c) Initial flux through coil C, \( \Phi_i = N_c \Phi_m = 2.121 \times 10^{-3} \text{ Wb} \)
Reversing the current in the solenoid reverses the flux through coil C, the magnitude remaining the same. But since the flux enters through the other face of the coil, the final flux through C is \( \Phi_f = -2.121 \times 10^{-3} \text{ Wb} \)
Therefore, the average emf induced in coil C,
\( e = - \frac{\Phi_f - \Phi_i}{\Delta t} = - \frac{(-2.121 - 2.121) \times 10^{-3}}{0.5} \)
\( = 2 \times 4.242 \times 10^{-3} = 8.484 \times 10^{-3} \text{ V} = 8.484 \text{ mV} \)
In simple words: We first find the magnetic field strength of the long coil, then how much of that field passes through the small coil. When we flip the electricity direction, the magnetic field flips too, creating a small pulse of 8.48 millivolts.

📝 Teacher's Note: When reversing current, the change in flux is \( 2\Phi \), not just \( \Phi \). This is a very common trap for students.

🎯 Exam Tip: For part (c), always show the step \( \Phi_f = -\Phi_i \). This demonstrates you understand the vector nature of magnetic flux.

Question 8. A search coil having 2000 turns with area \( 1.5 \text{ cm}^2 \) is placed in a magnetic field of 0.60 T. The coil is moved rapidly out of the field in a time of 0.2 second. Calculate the induced emf across the search coil.
Answer: Data: \( N = 2000, A_i = 1.5 \times 10^{-4} \text{ m}^2, A_f = 0, B = 0.6 \text{ T}, \Delta t = 0.2 \text{ s} \)
Initial flux, \( N\Phi_i = NBA_i = 2000(0.6)(1.5 \times 10^{-4}) = 0.18 \text{ Wb} \)
Final flux, \( N\Phi_f = 0 \)
Induced emf \( e = - \frac{\Delta (N\Phi)}{\Delta t} = \frac{0.18 - 0}{0.2} = 0.9 \text{ V} \)

= 0.18 Wb
Final flux, \( N\Phi_f = 0 \), since the coil is withdrawn out of the field.
Induced emf, \( e = -N \frac{\Delta \Phi_m}{\Delta t} = -N \frac{\Phi_f - \Phi_1}{\Delta t} \)
\( \therefore e = - \frac{0 - 0.18}{0.2} = 0.9\text{V} \)

Question 9. An aircraft of wing span of 50 m flies horizontally in earth’s magnetic field of \( 6 \times 10^{-5} \text{ T} \) at a speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft.
Answer:
Data: \( l = 50 \text{ m, } B = 6 \times 10^{-5} \text{ T, } v = 400 \text{ m/s} \)
The magnitude of the induced emf,
\( |e| = Blv = (6 \times 10^{-5})(400)(50) = 1.2 \text{V} \)
In simple words: When an airplane moves through the Earth's magnetic field, its wings act like a moving wire that cuts through magnetic lines, which creates a small electrical voltage between the two wingtips.

📝 Teacher's Note: Use the analogy of a comb moving through hair to explain "cutting" magnetic flux lines. Emphasize that the wings must be perpendicular to the magnetic field component for this simple formula to apply directly.

🎯 Exam Tip: Always state the formula \( e = Blv \) before substituting values. Ensure all units are in SI (meters, Tesla, and m/s) to get the final answer in Volts.

Question 10. A stiff semi-circular wire of radius R is rotated in a uniform magnetic field B about an axis passing through its ends. If the frequency of rotation of the wire is f, calculate the amplitude of the alternating emf induced in the wire.
Answer:
In one rotation, the wire traces out a circle of radius R, i.e., an area \( A = \pi R^2 \).
Therefore, the rate at which the wire traces out the area is
\( \frac{dA}{dt} = \text{frequency or rotation} \times A = fA \)
If the angle between the uniform magnetic field \( \vec{B} \) and the rotation axis is \( \theta \), the magnitude of the induced emf is
\( |e| = B \frac{dA}{dt} \cos \theta = BfA \cos \theta = Bf(\pi R^2) \cos \theta \)
so that the required amplitude is equal to \( Bf(\pi R^2) \).
In simple words: As the semi-circle spins, it sweeps through a circular area. Because it is moving through a magnetic field, the changing area it covers induces a pulse of electricity with every turn.

📝 Teacher's Note: Help students visualize that a rotating semi-circle effectively sweeps the same area as a full circle of the same radius over one complete period. This is why we use \( \pi R^2 \) instead of \( \frac{\pi R^2}{2} \) for the rate of change.

🎯 Exam Tip: The question asks for "amplitude," which means you are looking for the maximum value. Ensure your final expression is simplified and clearly labeled.

Question 11. Calculate the value of induced emf between the ends of an axle of a railway carriage 1.75 m long traveling on level ground with a uniform velocity of 50 km per hour. The vertical component of Earth’s magnetic field (\( B_v \)) is given to be \( 5 \times 10^{-5} \text{ T} \).
Answer:
Data: \( l = 1.75 \text{ m, } v = 50 \text{ km/h} = 50 \times \frac{5}{18} \text{ m/s} \).
\( B_v = 5 \times 10^{-5} \text{ T} \)
The area swept out by the wing per unit time = \( lv \).

\( \therefore \) The magnetic flux cut by the wing per unit time
\( = \frac{d\Phi_m}{dt} = B_v(lv) \)
\( = (5 \times 10^{-5})(1.75)(50 \times \frac{5}{18}) = 121.5 \times 10^{-5} \)
\( = 1.215 \text{ mWb/s} \)
Therefore, the magnitude of the induced emf,
\( |e| = 1.215 \text{ mV} \)
[Note: In the northern hemisphere, the vertical component of the Earth’s magnetic induction is downwards. Using Fleming’s right hand rule, the port (left) wing-tip would be positive.]
In simple words: A moving train axle cuts through the vertical magnetic lines of the Earth. This physical "cutting" action acts like a tiny battery, producing a very small voltage between the wheels.

📝 Teacher's Note: Point out the "wing" typo in the source text; in this context, it refers to the axle. Explain that horizontal motion only cuts the vertical component of the magnetic field.

🎯 Exam Tip: Conversion of units is key! Forgetting to convert km/h to m/s is the most common reason students lose marks on this specific problem.

Question 12. The value of mutual inductance of two coils is 10 mH. If the current in one of the coil changes from 5A to 1A in 0.2 s, calculate the value of emf induced in the other coil.
Answer:
Data: \( M = 10 \text{ mH} = 10^{-2} \text{ H, } I_{1i} = 5 \text{ A, } I_{1f} = 1 \text{ A,} \)
\( \Delta t = 0.2 \text{s} \)
The mutually induced emf in coil 2 due to the changing current in coil 1,
\( e_{21} = -M \frac{\Delta I_1}{\Delta t} = -M \frac{I_{1f} - I_{1i}}{\Delta t} \)
\( = -(10^{-2}) \left( \frac{1 - 5}{0.2} \right) = 0.2 \text{ V} \)
In simple words: When the electricity in the first coil drops suddenly, it creates a magnetic wave that "pushes" electricity into the second coil nearby, creating a voltage of 0.2 Volts.

📝 Teacher's Note: Use this to explain Lenz's Law. The negative sign indicates that the induced emf opposes the change in current that created it.

🎯 Exam Tip: When the current is decreasing, the change in current (\( \Delta I \)) is negative. This negative sign cancels with the formula's negative sign to give a positive emf magnitude.

Question 13. An emf of 96.0 mV is induced in the windings of a coil when the current in a nearby coil is increasing at the rate of 1.20 A/s. What is the mutual inductance (M) of the two coils?
Answer:
Data: \( |e_2| = 9.6 \times 10^{-2} \text{ V, } dI_1/dt = 1.2 \text{ A/s} \)
\( |e_2| = M \frac{dI_1}{dt} \)
Mutual inductance,
\( M = \frac{|e_2|}{dI_1/dt} = \frac{9.6 \times 10^{-2}}{1.2} \)

\( \implies 8 \times 10^{-2} \text{ H} \)
\( = 80 \text{ mH} \)
In simple words: This problem measures how efficiently two coils are linked together. By knowing the voltage produced by a certain change in current, we can calculate their "connection strength" or mutual inductance.

📝 Teacher's Note: Emphasize that mutual inductance \( M \) is constant for a fixed arrangement of two coils. It depends only on their geometry and the material between them.

🎯 Exam Tip: Pay attention to decimal places and powers of ten. \( 96.0 \text{ mV} \) must be correctly written as \( 0.096 \text{ V} \) for the formula to work.

Question 14. A long solenoid of length l, cross-sectional area A and having \( N_1 \) turns (primary coil) has a small coil of \( N_2 \) turns (secondary coil) wound about its centre. Determine the Mutual inductance (M) of the two coils.
Answer:
We assume the solenoid to be ideal and that all the flux from the solenoid passes through the outer coil C. For a steady current \( I_s \) through the solenoid, the uniform magnetic field inside the solenoid is
\( B = \mu_0 n I_s \dots \dots \dots (1) \)
Then, the magnetic flux through each turn of the coil due to the current in the solenoid is
\( \Phi_{CS} = BA = (\mu_0 n I_s)(\pi R^2) \dots \dots \dots (2) \)
Thus, their mutual inductance is
\( M = \frac{N \Phi_{CS}}{I_s} = \mu_0 \pi R^2 nN \dots \dots \dots (3) \)
Equation (2) is true as long as the magnetic field of the solenoid is entirely contained within the cross section of the coil C. Hence, M does not depend on the shape, size, or possible lack of close packing of the coil.
replacing n with \( N_1/l \) and N with \( N_2 \). \( M = \mu_0 A \frac{N_1 N_2}{l} \)
[Note: The answer given in the textbook misses out the factor of 1.]
In simple words: To find the mutual inductance, we look at how much magnetic field the first big coil makes inside itself and then calculate how much of that field is captured by the smaller coil wrapped around it.

📝 Teacher's Note: This is a standard derivation. Remind students that "n" represents the number of turns per unit length (\( N/l \)), while "N" is the total number of turns in the secondary coil.

🎯 Exam Tip: Clearly state the assumptions, such as the solenoid being "ideal" and "long," to justify using the uniform magnetic field formula.

Question 15. The primary and secondary coil of a transformer each have an inductance of \( 200 \times 10^{-6} \text{ H} \). The mutual inductance (M) between the windings is \( 4 \times 10^{-6} \text{ H} \). What percentage of the flux from one coil reaches the other?
Answer:
Data: \( L_P = L_S = 2 \times 10^{-4} \text{ H, } M = 4 \times 10^{-6} \text{ H} \)
\( M = K \sqrt{L_P L_S} \)
The coupling coefficient is
\( K = \frac{M}{\sqrt{L_P L_S}} = \frac{4 \times 10^{-6}}{\sqrt{(2 \times 10^{-4})^2}} = \frac{4 \times 10^{-6}}{2 \times 10^{-4}} \)

\( \implies 2 \times 10^{-2} \)
Therefore, the percentage of flux of the primary reaching the secondary is
\( 0.02 \times 100\% = 2\% \).
In simple words: The "coupling coefficient" tells us what fraction of the magnetic field from the first coil actually passes through the second coil. Here, only 2% of the magnetic energy is successfully shared.

📝 Teacher's Note: Explain that in an ideal transformer, the coupling coefficient \( K \) would be 1 (or 100%). A value of 0.02 indicates very poor magnetic coupling.

🎯 Exam Tip: Remember that "percentage of flux reaching the other" is simply another way of asking for the coupling coefficient expressed as a percentage.

Question 16. A toroidal ring, having 100 turns per cm of a thin wire is wound on a nonmagnetic metal rod of length 1 m and diameter 1 cm. If the permeability of bar is equal to that of free space (\( \mu_0 \)), calculate the magnetic field inside the bar (B) when the current (i) circulating through the turns is 1 A. Also determine the self-inductance
Answer:
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In simple words: A toroid is a doughnut-shaped coil. This question asks us to calculate how strong the magnetic field is inside that doughnut when electricity flows through the wire.

📝 Teacher's Note: For a toroid, the magnetic field is found using \( B = \mu_0 ni \). Ensure students convert "turns per cm" into the SI unit "turns per meter."

🎯 Exam Tip: When calculating self-inductance for a toroid, the formula \( L = \mu_0 n^2 A l \) is used, where \( l \) is the mean circumference of the toroid.