Electromagnetic Induction

JEE Main level, single correct, medium

94 of 153

Question 94 of 153

The figure shows a square loop LL of side 5cm5cm which is connected to a network of resistances. The whole setup is moving towards the right with a constant speed of 1cms−11cm{s}^{-1} . At some instant, a part of LL is in a uniform magnetic field of 1T1T perpendicular to the plane of the loop. If the resistance of LL is 1.7Ω,1.7\Omega , the current in the loop at that instant will be close to:

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Answer: (D) 170μA170\mu A

If BB is magnetic field, vv is velocity of rod and ll is the length of rod, then induced EMF in moving road is given by
ϵ=Bvl\epsilon =Bvl
=(1T)(1cms−1)(5cm)=\left(1T\right)(1 cm{s}^{-1})(5 cm)
=1×1×10−2×5×10−2=1\times 1\times {10}^{-2}\times 5\times {10}^{-2}
=5×10−4V=5\times {10}^{-4} V
As given circuit forms balanced whetstone bridge, the current through 3Ω3 \Omega is zero.
Equivalent resistance Req=4×24+2+1.7=3Ω{R}_{eq}=\dfrac{4\times 2}{4+2}+1.7=3\Omega
Current in the circuit i=ϵReq=5×10−43i=\dfrac{\epsilon }{{R}_{eq}}=\dfrac{5\times {10}^{-4}}{3}
=167μA≈170μA=167\mu A\approx 170\mu A

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