JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Electromagnetic Induction

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An emf of 0.08 V is induced in a metal rod of length 10 cm held normal to a uniform magnetic field of 0.4 T , when moves with a velocity of:

(A) $2 ms^{- 1}$ ✓ Correct
(B) $3.2 ms^{- 1}$
(C) $0.5 ms^{- 1}$
(D) $20 ms^{- 1}$

Answer: (A)

Solution

The motional emf across a rod of length

l

moving with speed

v

perpendicular to a field

B

ε = B l v

, since the rod is held normal to the field and sweeps the maximum flux. Rearranging for the speed gives

v = \frac{ε}{B l}

. Substituting

ε = 0.08 V

B = 0.4 T

and

l = 10 cm = 0.1 m

v = \frac{0.08}{0.4 \times 0.1} = \frac{0.08}{0.04}

. The rod moves with a velocity of

2 m s^{- 1}

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option

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