JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Electromagnetic Induction

A coil of inductance 1 H and resistance

100Ω

is connected to a battery of 6 V. Determine approximately :
(a) The time elapsed before the current acquires half of its steady - state value
(b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on. (Given

In 2 = 0.693

e^{- 3/2} = 0.25

)

(A) $t = 10 ms; U = 2 mJ$
(B) $t = 10 ms; U = 1 mJ$
(C) $t = 7 ms; U = 1 mJ$ ✓ Correct
(D) $t = 7 ms; U = 2 mJ$

Answer: (C)

Solution

For an R-L growth circuit,

i = \frac{E}{R} (1 - e^{- t / τ})

. When

i = \frac{E}{2 R}

, solving gives

t = τ ln 2 = \frac{L}{R} ln 2 = \frac{1}{100} \times 0.693 = 7 ms

. At

t = 15 ms

i = \frac{6}{100} (1 - e^{- 1.5}) = \frac{6}{100} (1 - 0.25) = 0.045 A

. Energy stored:

U = \frac{1}{2} L i^{2} = \frac{1}{2} (1) (0.045)^{2} \approx 1 mJ

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Electromagnetic Induction

Solution

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