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Question

(B) The motional electromotive force $(EMF)$ induced in a conductor of length $l$ moving with velocity $V$ in a magnetic field $B$ is given by $e = B

Vedclass · Accepted Answer

$e = \frac{1}{4}BLV$,$P$ is positive with respect to $Q$

Vedclass · Answer

(B) The motional electromotive force $(EMF)$ induced in a conductor of length $l$ moving with velocity $V$ in a magnetic field $B$ is given by $e = B l V \sin 	heta$,where $	heta$ is the angle between the velocity vector and the magnetic field. Here,the vertical segment of the loop containing the gap $PQ$ has an effective length $l = \frac{L}{4}$.As the loop moves to the right with velocity $V$ in a magnetic field $B$ directed into the paper,the motional $EMF$ induced in the segment of length $\frac{L}{4}$ is $e = B \left( \frac{L}{4} ight) V = \frac{1}{4}BLV$.Using Fleming's Right-Hand Rule or the Lorentz force law $(F = q(v 	imes B))$,the positive charges in the moving conductor are pushed towards the upper terminal $P$. Therefore,$P$ is at a higher potential than $Q$.

The loop shown moves with a constant velocity $V$ in a uniform magnetic field of magnetic induction $B$ directed into the paper. The potential difference between $P$ and $Q$ is:

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