JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Electromagnetic Induction

The time taken for the magnetic energy to reach

25%

of its maximum value, when solenoid of resistance R , inductance L is connected to a battery, is:

(A) $\frac{L}{R} ln (2)$ ✓ Correct
(B) $\frac{L}{R} ln (10)$
(C)Infinite
(D) $\frac{L}{R} ln (5)$

Answer: (A)

Solution

The current in an LR circuit grows as

i = i_{0} (1 - e^{- t / τ})

where

τ = L / R

and

i_{0} = V / R

. Magnetic energy

U = \frac{1}{2} L i^{2}

. Maximum energy is

U_{ma x} = \frac{1}{2} L i_{0}^{2}

. For

U = 0.25 U_{ma x}

, we have

\frac{1}{2} L i^{2} = 0.25 \cdot \frac{1}{2} L i_{0}^{2}

, giving

i = 0.5 i_{0}

. Thus

i_{0} (1 - e^{- t / τ}) = 0.5 i_{0}

, so

1 - e^{- t / τ} = 0.5

, giving

e^{- t / τ} = 0.5

, hence

t = τ ln (2) = \frac{L}{R} ln (2)

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option

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