Electromagnetic Induction

JEE Main level, single correct, medium

31 of 153

Question 31 of 153

An infinitely long straight wire carrying current I, one side opened rectangular loop and a conductor CC with a sliding connector are located in the same plane, as shown in the figure. The connector has length l. and resistance RR. It slides to the right with a velocity vv. The resistance of the conductor and the self inductance of the loop are negligible. The induced current in the loop, as a function of separation r, between the connector and the straight wire is

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Answer: (D) μ02πIvℓRr\dfrac{{\mu }_{0}}{2\pi }\dfrac{Iv\ell }{Rr}

The induced current in the wire will be due to the motional EMF developed in the moving resistor and that back emf developed at that point will be equal to VBLVBL
V=V=velocity of the conductor
B=B=perpendicular magnetic field
L=L=length of conductor
We know magnetic field due to a long wire is

B=μ0i2πrB=\dfrac{{\mu }_{0}i}{2\pi r}
Hence the induced current across wire will be :

I=eRI=\dfrac{e}{R}
I=μ02πIvℓRrI=\dfrac{{\mu }_{0}}{2\pi }\dfrac{Iv\ell }{Rr}

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