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MCQEasyJEE 2025Self & Mutual Inductance

JEE Physics 2025 Question with Solution

A solenoid having area AA and length ℓ\ell is filled with a material having relative permeability 22. The magnetic energy stored in the solenoid is:

  • A

    B2A2μ0\frac{B^2A}{2\mu_0}

  • B

    B2Aμ0\frac{B^2A}{\mu_0}

  • C

    B2A4μ0\frac{B^2A}{4\mu_0}

  • D

    B2A4μ0\frac{B^2A}{4\mu_0}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A solenoid has area AA, length ℓ\ell, and relative permeability μr=2\mu_r = 2.

Find: The magnetic energy stored in the solenoid.

The magnetic energy density in a medium is

u=B22μu = \frac{B^2}{2\mu}

where

μ=μrμ0\mu = \mu_r\mu_0

For the solenoid, the volume is

V=AℓV = A\ell

So the total magnetic energy is

U=uV=B22μ×Aℓ=B2Aℓ2μrμ0U = uV = \frac{B^2}{2\mu} \times A\ell = \frac{B^2A\ell}{2\mu_r\mu_0}

Substituting μr=2\mu_r = 2,

U=B2Aℓ2(2)μ0=B2Aℓ4μ0U = \frac{B^2A\ell}{2(2)\mu_0} = \frac{B^2A\ell}{4\mu_0}

The provided solution notes that the listed option corresponds to energy per unit length:

Uℓ=B2A4μ0\frac{U}{\ell} = \frac{B^2A}{4\mu_0}

Therefore, the correct option is D.

Option Discrepancy Note

The solution working gives total energy as

U=B2Aℓ4μ0U = \frac{B^2A\ell}{4\mu_0}

But the options are written without ℓ\ell. Hence the listed answer matches the expression for energy per unit length,

Uℓ=B2A4μ0\frac{U}{\ell} = \frac{B^2A}{4\mu_0}

the solution explicitly marks option D as correct, so that is taken as the final answer.

Common mistakes

  • Using μ=μ0\mu = \mu_0 instead of μ=μrμ0\mu = \mu_r\mu_0. This ignores the effect of the magnetic material. Always replace permeability by μrμ0\mu_r\mu_0 when a medium is present.

  • Confusing energy density with total energy. B22μ\frac{B^2}{2\mu} is energy per unit volume, not the total energy. Multiply by the solenoid volume AℓA\ell before comparing expressions.

  • Missing the discrepancy between total energy and energy per unit length. The derived total energy contains ℓ\ell, but the options do not. Use the solution conclusion and note that the marked option corresponds to Uℓ\frac{U}{\ell}.

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