Self Inductance — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Self Inductance MCQs with step-by-step solutions (21 questions). Part of Electromagnetic Induction. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Self Inductance · easy · theory
The self inductance of a coil is defined by the relation:
A. $L = \dfrac{\Phi}{NI}$
B. $L = \dfrac{N\Phi}{I}$ ✓ Correct
C. $L = \dfrac{I}{N\Phi}$
D. $L = N\Phi I$
Solution: It measures the flux linkage produced per unit current, and its SI unit is the henry.
Q2 — Self Inductance · easy · theory
The back EMF induced in a coil of self inductance $L$ when the current changes is:
A. $-LI$
B. $-\dfrac{L}{I}\dfrac{dt}{dI}$
C. $-\dfrac{dI}{Ldt}$
D. $-L\dfrac{dI}{dt}$ ✓ Correct
Solution: The minus sign shows the induced EMF opposes the change in current — the electrical analogue of inertia.
Q3 — Self Inductance · easy · theory
The energy stored in an inductor of inductance $L$ carrying current $I$ is:
A. $\dfrac{1}{2}LI$
B. $\dfrac{L}{2I^2}$
C. $LI^2$
D. $\dfrac{1}{2}LI^2$ ✓ Correct
Solution: This energy resides in the magnetic field and is returned to the circuit when the current falls.
Q4 — Self Inductance · medium · theory
The self inductance of a long solenoid of $N$ turns is proportional to:
A. $N$
B. $N^2$ ✓ Correct
C. $\sqrt{N}$
D. $\dfrac{1}{N}$
Solution: $L = \dfrac{\mu_0N^2A}{l}$: each turn both produces and links the flux, giving the square.
Q5 — Self Inductance · medium · theory
The self inductance of a coil depends on:
A. The current flowing through it
B. The resistance of the wire
C. Its geometry and the magnetic properties of its core ✓ Correct
D. The EMF applied to it
Solution: Like capacitance in electrostatics, inductance is fixed by construction rather than by the current carried.
Q6 — Self Inductance · medium · theory
Self induction is sometimes called the inertia of electricity because it:
A. Dissipates energy as heat
B. Opposes any change in the current through the circuit ✓ Correct
C. Increases the resistance of the circuit
D. Stores charge like a capacitor
Solution: Just as mass resists a change of velocity, inductance resists a change of current.
Q7 — Self Inductance · easy · theory
Inserting a soft iron core into a solenoid causes its self inductance to:
A. Become zero
B. Increase greatly ✓ Correct
C. Remain unchanged
D. Decrease greatly
Solution: $L = \dfrac{\mu_r\mu_0N^2A}{l}$, and $\mu_r$ for soft iron is of the order of thousands.
Q8 — Self Inductance · medium · theory
The SI unit of self inductance is the henry, which is equivalent to:
A. $\text{Wb}/\text{s}$
B. $\text{V}\cdot\text{A}/\text{s}$
C. $\text{V}\cdot\text{s}/\text{A}$ ✓ Correct
D. $\text{A}\cdot\text{s}/\text{V}$
Solution: From $e = L\dfrac{dI}{dt}$, $L$ has units of volt second per ampere, equivalently weber per ampere.
Q9 — Self Inductance · easy · numerical
An inductor of $5\text{ H}$ carries a steady current of $2\text{ A}$. The energy stored in it is:
A. $2.5\text{ J}$
B. $20\text{ J}$
C. $10\text{ J}$ ✓ Correct
D. $5\text{ J}$
Solution: $U = \dfrac{1}{2}LI^2 = \dfrac{1}{2}(5)(4) = 10\text{ J}$.
Q10 — Self Inductance · easy · numerical
The current in a $2\text{ H}$ inductor changes at $5\text{ A}/\text{s}$. The back EMF is:
A. $0.4\text{ V}$
B. $20\text{ V}$
C. $2.5\text{ V}$
D. $10\text{ V}$ ✓ Correct
Solution: $|e| = L\dfrac{dI}{dt} = 2 \times 5 = 10\text{ V}$.
Q11 — Self Inductance · medium · numerical
The current in a $0.5\text{ H}$ inductor rises from zero to $4\text{ A}$ in $0.2\text{ s}$. The induced EMF is:
A. $20\text{ V}$
B. $1\text{ V}$
C. $10\text{ V}$ ✓ Correct
D. $2\text{ V}$
Solution: $|e| = L\dfrac{\Delta I}{\Delta t} = 0.5 \times \dfrac{4}{0.2} = 0.5 \times 20 = 10\text{ V}$.
Q12 — Self Inductance · easy · numerical
An inductor of $4\text{ H}$ carries a current of $3\text{ A}$. The energy stored is:
A. $6\text{ J}$
B. $18\text{ J}$ ✓ Correct
C. $12\text{ J}$
D. $36\text{ J}$
Solution: $U = \dfrac{1}{2}(4)(9) = 18\text{ J}$.
Q13 — Self Inductance · medium · numerical
If the number of turns of a solenoid is doubled while its length and area stay the same, its self inductance becomes:
A. Twice as large
B. Half as large
C. Unchanged
D. Four times as large ✓ Correct
Solution: $L \propto N^2$, so doubling $N$ multiplies $L$ by four.
Q14 — Self Inductance · hard · numerical
A solenoid of $1000$ turns, area $10^{-4}\text{ m}^2$ and length $0.5\text{ m}$ has self inductance approximately:
A. $2.5 \times 10^{-6}\text{ H}$
B. $2.5 \times 10^{-4}\text{ H}$ ✓ Correct
C. $1.26 \times 10^{-4}\text{ H}$
D. $2.5 \times 10^{-2}\text{ H}$
Solution: $L = \dfrac{\mu_0N^2A}{l} = \dfrac{4\pi \times 10^{-7} \times 10^6 \times 10^{-4}}{0.5} \approx 2.5 \times 10^{-4}\text{ H}$.
Q15 — Self Inductance · medium · numerical
A coil of $200$ turns has a flux of $0.01\text{ Wb}$ linked with it when carrying $4\text{ A}$. Its self inductance is:
A. $0.5\text{ H}$ ✓ Correct
B. $2\text{ H}$
C. $5\text{ H}$
D. $0.05\text{ H}$
Solution: $L = \dfrac{N\Phi}{I} = \dfrac{200 \times 0.01}{4} = 0.5\text{ H}$.
Q16 — Self Inductance · medium · numerical
An inductor of $0.2\text{ H}$ carries a current of $10\text{ A}$. The energy stored is:
A. $10\text{ J}$ ✓ Correct
B. $20\text{ J}$
C. $1\text{ J}$
D. $2\text{ J}$
Solution: $U = \dfrac{1}{2}(0.2)(100) = 10\text{ J}$.
Q17 — Self Inductance · easy · numerical
If the current through an inductor is doubled, the energy stored becomes:
A. Twice as large
B. Half as large
C. Four times as large ✓ Correct
D. Unchanged
Solution: $U = \dfrac{1}{2}LI^2 \propto I^2$.
Q18 — Self Inductance · medium · numerical
A back EMF of $20\text{ V}$ appears when the current in a coil changes at $4\text{ A}/\text{s}$. The self inductance is:
A. $24\text{ H}$
B. $5\text{ H}$ ✓ Correct
C. $0.2\text{ H}$
D. $80\text{ H}$
Solution: $L = \dfrac{e}{dI/dt} = \dfrac{20}{4} = 5\text{ H}$.
Q19 — Self Inductance · medium · numerical
A solenoid is fitted with a core of relative permeability $500$. Its self inductance becomes:
A. $500$ times smaller
B. Unchanged
C. $250000$ times as large
D. $500$ times as large ✓ Correct
Solution: $L \propto \mu_r$, so the core multiplies the inductance by its relative permeability.
Q20 — Self Inductance · hard · numerical
An inductor of $10\text{ mH}$ carries a current of $0.5\text{ A}$. The energy stored is:
A. $2.5 \times 10^{-3}\text{ J}$
B. $1.25 \times 10^{-2}\text{ J}$
C. $5 \times 10^{-3}\text{ J}$
D. $1.25 \times 10^{-3}\text{ J}$ ✓ Correct
Solution: $U = \dfrac{1}{2}(10^{-2})(0.25) = 1.25 \times 10^{-3}\text{ J}$.
Q21 — Self Inductance · medium · numerical
A back EMF of $100\text{ V}$ is required from a $0.25\text{ H}$ inductor. The rate of change of current must be:
A. $400\text{ A}/\text{s}$ ✓ Correct
B. $4\text{ A}/\text{s}$
C. $25\text{ A}/\text{s}$
D. $40\text{ A}/\text{s}$
Solution: $\dfrac{dI}{dt} = \dfrac{e}{L} = \dfrac{100}{0.25} = 400\text{ A}/\text{s}$.