JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

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 2 $Ω$ 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
$7.5 \times 1 0^{- 3}$

Solution

Motional EMF: $ε = B v ℓ = 0.1 \times 1.5 \times 1 = 0.15$ V.

Induced current: $i = ε / R = 0.15/2 = 0.075$ A.

Magnetic force on the current-carrying rod opposes the motion: $F = B i ℓ = 0.1 \times 0.075 \times 1 = 7.5 \times 1 0^{- 3}$ N. The external force must equal this in magnitude to keep the rod moving at constant speed.

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