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PhysicsJEE Main 22 Jan 2026, Shift 1·Q25·Numerical·Moderate

Electromagnetic Induction · Inductance

Inductance of a coil with 10410^{4} turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of 10 Ω10\ \Omega. The energy density in the inductor when the current reaches (1e)\left(\dfrac{1}{e}\right) of its maximum value is απ×1e2 J/m3\alpha\pi\times\dfrac{1}{e^2}\ \mathrm{J/m^3}. The value of α\alpha is . (μ0=4π×10−7\mu_0 = 4\pi\times10^{-7} Tm/A). (Take the length of the coil as 1 m)
We corrected a printing error in the original paper

Assumption “Take the length of the coil as 1 m” added; the source gives neither the length nor the turns per unit length, and the key 20 needs n=104n = 10^4 turns/m (consistent with L=10L = 10 mH for a long solenoid).

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