Electromagnetic Induction

JEE Main level, numerical answer, medium

86 of 153

Question 86 of 153

A 1m1m long metal rod XYXY completes the circuit as shown in figure. The plane of the circuit is perpendicular to the magnetic field of flux density 0.15T0.15T. If the resistance of the circuit is 5Ω5Ω, the force needed to move the rod in direction, as indicated, with a constant speed of 4ms−14m{s}^{-1} will be _______10–3N{10}^{–3}N.

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Answer: 18

The force on the rod is given by
F=ilBF=ilB
⇒F=BvlR×lB\Rightarrow F=\dfrac{Bvl}{R}\times lB
⇒F=B2l2vR...(i)\Rightarrow F=\dfrac{{B}^{2}{l}^{2}v}{R}...(i)
The data given is
R=5Ωl=1mv=4ms−1B=0.15TR=5\Omega l=1m v=4m{s}^{-1} B=0.15T
Substituting the values in equation (i)
F=(0.15)2×(1)2×45F=\dfrac{{\left(0.15\right)}^{2}\times {\left(1\right)}^{2}\times 4}{5}
=18×10−3N=18\times {10}^{-3}N

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