The force on the rod is given by
F=ilB
⇒F=RBvl×lB
⇒F=RB2l2v...(i)
The data given is
R=5Ωl=1mv=4ms−1B=0.15T
Substituting the values in equation (i)
F=5(0.15)2×(1)2×4
=18×10−3N
JEE Main 2023 — Physics Electromagnetism
A 1m 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 0.15T. If the resistance of the circuit is 5Ω, the force needed to move the rod in direction, as indicated, with a constant speed of 4ms−1 will be _______10–3N.

Held on 10 Apr 2023 · Verified 6 Jul 2026.
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