Self and Mutual Inductances — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Self and Mutual Inductances MCQs with step-by-step solutions (18 questions). Part of Electromagnetic Induction. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Self and Mutual Inductances · medium · theory
A light bulb and an inductor coil are connected to an AC source through a key, as shown in the figure. The key is closed and after some time an iron rod is inserted into the interior of the inductor. The glow of the light bulb
A. decreases ✓ Correct
B. remains unchanged
C. will fluctuate
D. increases
Solution: Inserting an iron core increases the inductance $L$ of the coil, which increases the inductive reactance $X_L=\omega L$. Since the circuit current is $I=e/X_L$, the current decreases, so the bulb's glow decreases.
Q2 — Self and Mutual Inductances · medium · numerical
A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is $4\times10^{-3}$ Wb. The self-inductance of the solenoid is
A. $3\text{ H}$
B. $2\text{ H}$
C. $1\text{ H}$ ✓ Correct
D. $4\text{ H}$
Solution: $L=\dfrac{N\phi}{I}=\dfrac{1000\times4\times10^{-3}}{4}=1\text{ H}$.
Q3 — Self and Mutual Inductances · medium · numerical
A long solenoid has 500 turns. When a current of 2 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4\times10^{-3}$ Wb. The self-inductance of the solenoid is
A. $2.5\text{ H}$
B. $2\text{ H}$
C. $1\text{ H}$ ✓ Correct
D. $4\text{ H}$
Solution: Net flux linkage, $\phi_{net}=N\phi=500\times4\times10^{-3}=2\text{ Wb}$. Self-inductance $L=\dfrac{\phi_{net}}{i}=\dfrac{2}{2}=1\text{ H}$.
Q4 — Self and Mutual Inductances · medium · numerical
A varying current in a coil changes from $10\,\text{A}$ to zero in $0.5\,\text{s}$. If the average emf induced in the coil is $220\,\text{V}$, the self-inductance of the coil is
A. $5\,\text{H}$
B. $6\,\text{H}$
C. $11\,\text{H}$ ✓ Correct
D. $12\,\text{H}$
Solution: $e=L\dfrac{di}{dt} \Rightarrow 220=L\times\dfrac{10}{0.5} \Rightarrow L=\dfrac{220\times0.5}{10}=11\,\text{H}$.
Q5 — Self and Mutual Inductances · medium · theory
If $N$ is the number of turns in a coil, the value of self-inductance varies as
A. $N^0$
B. $N$
C. $N^2$ ✓ Correct
D. $N^{-2}$
Solution: Magnetic flux $\phi=BA$ and the field at the centre of a circular coil is $B=\dfrac{\mu_0Ni}{2R}$. The total flux linkage is $N\phi=\dfrac{\mu_0N^2iA}{2R}$, so $L=\dfrac{N\phi}{i}=\dfrac{\mu_0N^2A}{2R}$. Hence $L\propto N^2$.
Q6 — Self and Mutual Inductances · medium · numerical
What is the self-inductance of a coil which produces $5\,\text{V}$ when the current changes from $3\,\text{A}$ to $2\,\text{A}$ in one millisecond?
A. $5000\,\text{H}$
B. $5\,\text{mH}$ ✓ Correct
C. $50\,\text{H}$
D. $5\,\text{H}$
Solution: $|e|=L\dfrac{di}{dt} \Rightarrow L=\dfrac{|e|}{di/dt}=\dfrac{5}{1/(1\times10^{-3})}=5\times10^{-3}\,\text{H}=5\,\text{mH}$.
Q7 — Self and Mutual Inductances · medium · numerical
A $100\,\text{mH}$ coil carries a current of $1\,\text{A}$. Energy stored in its magnetic field is
A. $0.5\,\text{J}$
B. $1\,\text{A}$
C. $0.05\,\text{J}$ ✓ Correct
D. $0.1\,\text{J}$
Solution: $E=\dfrac{1}{2}Li^2=\dfrac{1}{2}\times(100\times10^{-3})\times1^2=0.05\,\text{J}$.
Q8 — Self and Mutual Inductances · medium · theory
If the number of turns per unit length of a coil of a solenoid is doubled, the self-inductance of the solenoid will
A. remain unchanged
B. be halved
C. be doubled
D. become four times ✓ Correct
Solution: For a long solenoid, $L=\mu_0n^2Al$, where $n$ is the number of turns per unit length. So $L\propto n^2$; doubling $n$ makes $L$ four times its original value.
Q9 — Self and Mutual Inductances · medium · theory
An inductor may store energy in
A. its electric field
B. its coils
C. its magnetic field ✓ Correct
D. Both in electric and magnetic fields
Solution: The induced emf is $e=-L\dfrac{di}{dt}$, and the work done in establishing a current $i_0$ is $W=\displaystyle\int_0^{i_0}Li\,di=\dfrac{1}{2}Li_0^2$. This work is stored as energy in the inductor's magnetic field.
Q10 — Self and Mutual Inductances · medium · theory
Energy in a current-carrying coil is stored in the form of
A. electric field
B. magnetic field ✓ Correct
C. dielectric strength
D. heat
Solution: The energy stored in an inductor carrying current $i_0$ is $W=\dfrac{1}{2}Li_0^2$, and this energy resides in the magnetic field set up by the current.
Q11 — Self and Mutual Inductances · medium · numerical
A long solenoid of diameter 0.1 m has $2\times10^4$ turns per metre. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is $10\pi^2\ \Omega$, the total charge flowing through the coil during this time is
A. $32\pi\ \mu\text{C}$
B. $16\pi\ \mu\text{C}$
C. $32\ \mu\text{C}$ ✓ Correct
D. $16\ \mu\text{C}$
Solution: Charge $q=\dfrac{\Delta\phi}{R}=\dfrac{N_2\mu_0n_1\Delta I\,\pi r^2}{R}$, with $n_1=2\times10^4\ \text{m}^{-1}$, $N_2=100$, $r=0.01\text{ m}$, $\Delta I=4\text{ A}$, $R=10\pi^2\ \Omega$. $q=\dfrac{100\times4\pi\times10^{-7}\times2\times10^4\times4\times\pi\times10^{-4}}{10\pi^2}=32\times10^{-6}\text{ C}=32\ \mu\text{C}$.
Q12 — Self and Mutual Inductances · medium · numerical
Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres coinciding. If $R_1\gg R_2$, the mutual inductance $M$ between them will be directly proportional to
A. $\dfrac{R_1}{R_2}$
B. $\dfrac{R_2}{R_1}$
C. $\dfrac{R_1^2}{R_2}$
D. $\dfrac{R_2^2}{R_1}$ ✓ Correct
Solution: The magnetic field at the common centre due to current $i$ in the outer loop (radius $R_1$) is $B=\dfrac{\mu_0i}{2R_1}$. Flux through the inner loop (area $\pi R_2^2$) is $\phi=BA=\dfrac{\mu_0i}{2R_1}\pi R_2^2$. Since $\phi=Mi$, $M=\dfrac{\mu_0\pi R_2^2}{2R_1}\propto\dfrac{R_2^2}{R_1}$.
Q13 — Self and Mutual Inductances · medium · numerical
The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance
A. $1.389\text{ H}$
B. $138.88\text{ H}$
C. $0.138\text{ H}$
D. $13.89\text{ H}$ ✓ Correct
Solution: Energy stored, $U=\dfrac{1}{2}LI_0^2\Rightarrow L=\dfrac{2U}{I_0^2}=\dfrac{2\times25\times10^{-3}}{(60\times10^{-3})^2}=\dfrac{0.05}{3.6\times10^{-3}}=13.89\text{ H}$.
Q14 — Self and Mutual Inductances · medium · numerical
Two coils of self-inductances $2\,\text{mH}$ and $8\,\text{mH}$ are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is
A. $10\,\text{mH}$
B. $6\,\text{mH}$
C. $4\,\text{mH}$ ✓ Correct
D. $16\,\text{mH}$
Solution: When the flux linkage between the two coils is maximum (coupling coefficient $k=1$), the mutual inductance is $M=\sqrt{L_1L_2}=\sqrt{2\times8}=\sqrt{16}=4\,\text{mH}$.
Q15 — Self and Mutual Inductances · medium · numerical
In an inductor of self-inductance $L=2\,\text{mH}$, current changes with time according to the relation $i=t^2e^{-t}$. At what time is the emf zero?
A. $4\,\text{s}$
B. $3\,\text{s}$
C. $2\,\text{s}$ ✓ Correct
D. $1\,\text{s}$
Solution: $e=-L\dfrac{di}{dt}=0 \Rightarrow \dfrac{di}{dt}=0$. Now $\dfrac{d}{dt}(t^2e^{-t})=2te^{-t}-t^2e^{-t}=te^{-t}(2-t)=0$. Since $t=0$ is trivial (and $te^{-t}\neq0$ for $t>0$), $t=2\,\text{s}$.
Q16 — Self and Mutual Inductances · medium · numerical
Two coils have a mutual inductance of $0.005\,\text{H}$. The current changes in the first coil according to the equation $i=i_0\sin\omega t$, where $i_0=10\,\text{A}$ and $\omega=100\pi\,\text{rad/s}$. The maximum value of emf in the second coil is
A. $2\pi\,\text{V}$
B. $5\pi\,\text{V}$ ✓ Correct
C. $\pi\,\text{V}$
D. $4\pi\,\text{V}$
Solution: $e=M\dfrac{di}{dt}=Mi_0\omega\cos\omega t$. The maximum emf is $e_{max}=Mi_0\omega=0.005\times10\times100\pi=5\pi\,\text{V}$.
Q17 — Self and Mutual Inductances · medium · numerical
The current in a self-inductance $L=40\,\text{mH}$ is to be increased uniformly from $1\,\text{A}$ to $11\,\text{A}$ in $4$ milliseconds. The emf induced in the inductor during the process is
A. $100\,\text{V}$ ✓ Correct
B. $0.4\,\text{V}$
C. $4\,\text{V}$
D. $440\,\text{V}$
Solution: $e=L\dfrac{di}{dt}=40\times10^{-3}\times\dfrac{11-1}{4\times10^{-3}}=40\times10^{-3}\times2500=100\,\text{V}$.
Q18 — Self and Mutual Inductances · medium · theory
In the circuit shown, an inductor $L$ and a bulb $B$ are connected in series with a battery and a switch $K$. The bulb will suddenly become bright if
A. contact is made or broken
B. contact is made
C. contact is broken ✓ Correct
D. None of the above
Solution: When the circuit contact is suddenly broken, the rapidly collapsing current in the inductor induces a self-induced emf that drives a momentary current in the same direction as the original current, so the bulb $B$ flashes suddenly bright before going out.