Mutual Induction - NEET Physics Questions
Question 1: moderate

The mutual inductance between a primary and secondary circuits is 0.5 H. The resistance of the
primary and the secondary circuits are 20Ω and 5Ω respectively. To generate a current of 0.4 A
in the secondary, current in the primary must be changed at the rate of :

1. 4.0 A/s
2. 1.6 A/s
3. 16.0 A/s
4. 8.0 A/s
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Question 2: difficult

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm² and length = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is
\[\left( \mu=4\pi\times 10^{-7} T m A^{-1}\right)\] :

1. \[4.8\pi\times 10^{-4} H\]
2. \[4.8\pi\times 10^{-5} H\]
3. \[2.4\pi\times 10^{-4} H\]
4. \[2.4\pi\times 10^{-5} H\]
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Question 3: difficult

What is the mutual inductance of a two-loop system as shown with centre separation l ?

1. \[\frac{\mu_{0}\pi a^{4}}{8l^{3}}\]
2. \[\frac{\mu_{0}\pi a^{4}}{4l^{3}}\]
3. \[\frac{\mu_{0}\pi a^{4}}{6l^{3}}\]
4. \[\frac{\mu_{0}\pi a^{4}}{2l^{3}}\]
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Question 4: easy

Two circular coils can be arranged in any of the three situations shown in the figure. Their
mutual inductance will be :

1. maximum In situation (a)
2. maximum In situation (b)
3. maximum In situation (c)
4. the same in all situations
View Answer

Based on the visual arrangement of the coils, the mutual inductance is maximum in situation (a).

This is because the coils are placed co-axially (one above the other), allowing the maximum amount of magnetic flux from the primary coil to pass through the secondary coil, resulting in the highest coupling coefficient.

Question 5: easy

Two coils X and Y are placed in a circuit such that a current changes by 2 A in coil X and the magnetic flux change of 0.4 Wb occurs in coil Y. The value of mutual inductance of coils is :

1. 0.2 H
2. 2 H
3. 0.5 H
4. 5 H
View Answer

The mutual inductance ($M$) is calculated by the ratio of flux change in coil Y to the current change in coil X:

$$M = \frac{\Delta \phi_Y}{\Delta I_X} = \frac{0.4 \text{ Wb}}{2 \text{ A}}$$

$M = 0.2 \text{ H}$

Question 6: easy

A small square loop of wire of side l is placed inside a large square loop of wire of side L ( L > l ). The loops are coplanar and their centres coincide. The mutual inductance of the system is proportional to :

1. l / L
2. / L
3. L/l
4. L²/l
View Answer

The mutual inductance M is found by calculating the magnetic flux through the small loop due to the current I in the large loop. Using the formula for the magnetic field at the center of a square loop, B  \frac{I}{L}.

Since the small loop is much smaller than the large one \(L \gg l)\, the field is approximately uniform across its area \A = l^2\. The flux \Phi = B \cdot A$ is therefore proportional to \\frac{I}{L} \cdot l^2\.

Because $M = \frac{\Phi}{I}$, the mutual inductance scales as:

$$M \propto \frac{l^2}{L}$$