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NVAEasyJEE 2023Self & Mutual Inductance

JEE Physics 2023 Question with Solution

A coil has an inductance of 2 H2 \, \text{H} and resistance of 4 Ω4 \, \Omega. A 10 V10 \, \text{V} is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be _____ ×10−2\times 10^{-2} J\text{J}.

Answer

Correct answer:625

Step-by-step solution

Standard Method

Given:

  • Inductance L=2 HL = 2 \, \text{H}
  • Resistance R=4 ΩR = 4 \, \Omega
  • Applied voltage V=10 VV = 10 \, \text{V}

Find: The energy stored in the magnetic field when the current reaches equilibrium.

At steady state, the inductor behaves as a short circuit, so the circuit reduces to a resistor connected across the source.

Circuit showing a 10 V source, 2 H inductor and 4 ohm resistor in series, then equivalent steady state circuit with only 4 ohm resistor and 10 V source.

Using Ohm's law, the steady-state current is

I=VRI = \frac{V}{R}

Substituting the values,

I=104=52=2.5 AI = \frac{10}{4} = \frac{5}{2} = 2.5 \, \text{A}

The energy stored in the inductor is

E=12LI2E = \frac{1}{2}LI^2

Substituting L=2 HL = 2 \, \text{H} and I=2.5 AI = 2.5 \, \text{A},

E=12⋅2⋅(2.5)2E = \frac{1}{2} \cdot 2 \cdot (2.5)^2 E=6.25 JE = 6.25 \, \text{J}

Now express it in the required form:

E=625×10−2 JE = 625 \times 10^{-2} \, \text{J}

Therefore, the required numerical value is 625.

Expanded Working

Given: L=2 HL = 2 \, \text{H}, R=4 ΩR = 4 \, \Omega, V=10 VV = 10 \, \text{V}

Find: The blank in _____ \times 10^{-2} \, \text{J}.

When sufficient time has passed after applying a DC source, the current becomes constant. Then

didt=0\frac{di}{dt} = 0

and the induced emf across the inductor becomes zero. Hence the inductor offers no opposition in steady state and acts like a wire.

So the equilibrium current is determined only by the resistance:

I=VR=104=2.5 AI = \frac{V}{R} = \frac{10}{4} = 2.5 \, \text{A}

Magnetic energy stored in an inductor is

E=12LI2E = \frac{1}{2}LI^2

Therefore,

E=12(2)(2.5)2E = \frac{1}{2}(2)(2.5)^2 E=1⋅6.25=6.25 JE = 1 \cdot 6.25 = 6.25 \, \text{J}

To match the asked format,

6.25 J=625×10−2 J6.25 \, \text{J} = 625 \times 10^{-2} \, \text{J}

Hence the answer is 625.

Common mistakes

  • Using the transient current instead of the equilibrium current is incorrect because the question asks for the value after the current has fully built up. First take the steady-state current I=VRI = \frac{V}{R}, then compute energy.

  • Treating the inductor as having resistance LL is wrong because inductance and resistance are different quantities. At steady state for DC, the inductor behaves like a short circuit, not like a resistor of 2 Ω2 \, \Omega.

  • Forgetting the factor 12\frac{1}{2} in E=12LI2E = \frac{1}{2}LI^2 gives an incorrect energy. Always use the full magnetic energy expression before substituting values.

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