JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Electromagnetic Induction

A coil has an inductance of 2 H and resistance of

4Ω

. A 10 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

____

\times 1 0^{- 2}

Answer: 625

Solution

At equilibrium, the current reaches its steady-state value

I = \frac{V}{R} = \frac{10}{4} = 2.5

A. The energy stored in the inductor is

E = \frac{1}{2} L I^{2} = \frac{1}{2} \times 2 \times (2.5)^{2} = \frac{1}{2} \times 2 \times 6.25 = 6.25

= 625 \times 1 0^{- 2}

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Integer answer

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