Electromagnetic Induction

JEE Main level, single correct, medium

73 of 153

Question 73 of 153

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 2Ω2 \Omega then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s1.5 \mathrm{~m} / \mathrm{s} is ____\_\_\_\_ N.

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Answer: (C) 7.5×10−37.5 \times 10^{-3}

Given: l=1l = 1 m, B=0.10B = 0.10 T, R=2R = 2 Ω, v=1.5v = 1.5 m/s
Motional EMF: ε=Blv=0.10×1×1.5=0.15\varepsilon = Blv = 0.10 \times 1 \times 1.5 = 0.15 V
Current: I=εR=0.152=0.075I = \frac{\varepsilon}{R} = \frac{0.15}{2} = 0.075 A
Magnetic force on rod: F=BIl=0.10×0.075×1=7.5×10−3F = BIl = 0.10 \times 0.075 \times 1 = 7.5 \times 10^{-3} N
For constant velocity, applied force equals magnetic force.

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