JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Electromagnetic Induction

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A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is

2Ω

then the force needed to move the rod towards right with constant speed (v) of

1.5 m / s

is ______ N.

(A) $7.5 \times 1 0^{- 2}$
(B) $5.7 \times 1 0^{- 3}$
(C) $5.7 \times 1 0^{- 2}$
(D) $7.5 \times 1 0^{- 3}$ ✓ Correct

Answer: (D)

Solution

The moving rod generates an emf

ε = B l v

and a current

i = \frac{B l v}{R}

. To keep the rod at constant speed the applied force must balance the magnetic force on it,

F = B i l = \frac{B ^{2} l ^{2} v}{R}

. Substituting,

F = \frac{( 0.10 ) ^{2} \times 1 ^{2} \times 1.5}{2} = 7.5 \times 1 0^{- 3} N

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option

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