JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Electromagnetic Induction

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A 1 m long metal rod XY completes the circuit as shown in figure. The plane of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the circuit is

5Ω

, the force needed to move the rod in direction, as indicated, with a constant speed of

4 m / s

will be

____

1 0^{- 3} N

Answer: 18

Solution

The motional emf is

ε = B ℓ v

, driving a current

i = \frac{B ℓ v}{R}

through the rod. To keep the rod moving at constant speed the applied force must balance the magnetic force

F = B i ℓ

, giving

F = \frac{B ^{2} ℓ ^{2} v}{R}

. Substituting

B = 0.15

ℓ = 1

v = 4 m s^{- 1}

and

R = 5 Ω

F = \frac{( 0.15 ) ^{2} ( 1 ) ^{2} ( 4 )}{5} = \frac{0.09}{5} = 0.018

N. The required force is

18 \times 1 0^{- 3}

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Integer answer

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