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MCQEasyJEE 2023Self & Mutual Inductance

JEE Physics 2023 Question with Solution

A 12 V12 \, \text{V} battery connected to a coil of resistance 6 Ω6\,\Omega through a switch drives a constant current in the circuit. The switch is opened in 1 ms1\,ms. The emf induced across the coil is 20 V20 \, \text{V}. The inductance of the coil is

  • A

    5 mH5\,mH

  • B

    8 mH8\,mH

  • C

    10 mH10\,mH

  • D

    12 mH12\,mH

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Battery voltage 12 V12 \, \text{V}, resistance 6 Ω6\,\Omega, induced emf 20 V20 \, \text{V}, and switching time 1 ms=10−3 s1\,ms = 10^{-3}\,s.

Find: The inductance of the coil.

When the current is steady, the inductive emf is zero. Hence,

I=VR=126=2 AI = \frac{V}{R} = \frac{12}{6} = 2 \, \text{A}

For an inductor, the induced emf is

E=L ΔIΔt\mathcal{E} = L\,\frac{\Delta I}{\Delta t}

Substitute the given values:

E=20 V,ΔI=2 A,Δt=10−3 s\mathcal{E} = 20 \, \text{V}, \quad \Delta I = 2 \, \text{A}, \quad \Delta t = 10^{-3} \, \text{s}

So,

20=L×210−320 = L \times \frac{2}{10^{-3}}

Therefore,

L=20×10−32=10×10−3 HL = \frac{20 \times 10^{-3}}{2} = 10 \times 10^{-3} \, \text{H}

Thus,

L=10 mHL = 10\,mH

Therefore, the inductance of the coil is 10 mH10\,mH and the correct option is C.

Use steady current directly

Given: The current was steady before opening the switch, so first use the resistor relation.

Find: The inductance LL.

Since the battery and resistance are known,

I=126=2 AI = \frac{12}{6} = 2 \, \text{A}

Now apply

L=E ΔtΔIL = \frac{\mathcal{E}\,\Delta t}{\Delta I}

Substituting,

L=20×10−32=10×10−3 H=10 mHL = \frac{20 \times 10^{-3}}{2} = 10 \times 10^{-3} \, \text{H} = 10\,mH

This works because the current falls from its steady value to zero when the switch is opened. Therefore, the correct option is C.

Common mistakes

  • Using the battery voltage 12 V12 \, \text{V} as the induced emf is incorrect because the question separately gives the induced emf across the coil as 20 V20 \, \text{V}. Use the induced emf value in E=L ΔIΔt\mathcal{E} = L\,\frac{\Delta I}{\Delta t}.

  • Forgetting to find the steady current before opening the switch is wrong because the change in current is based on the initial current in the circuit. First compute I=VRI = \frac{V}{R}, then use ΔI=2 A\Delta I = 2 \, \text{A}.

  • Not converting 1 ms1\,ms into 10−3 s10^{-3}\,s gives an incorrect inductance by a factor of 10001000. Always convert milliseconds to seconds before substitution.

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