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Byju's Answer

Standard XII

Physics

Motional EMF

The loop show...

Question

The loop shown moves with a velocity $v$ in a uniform magnetic field of magnitude $B$ , directed into the paper. The potential difference between points $P$ and $Q$ is $e$ . Then

e = \frac{1}{2} B l v

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e = B l v

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P

is positive with respect to

Q

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Q

is positive with respect to

P

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Solution

The correct options are
A $e = \frac{1}{2} B l v$
C $P$ is positive with respect to $Q$
Potential difference across a rod of length $l$ travelling perpendicular to magnetic field is $B v l$ .
Here potential difference across the side of square opposite to that containing PQ is $B v L$ .
Potential difference between point P and the corner of its side+Potential difference between point Q and the corner of its side= $\frac{1}{2} B v L$ .
Thus the potential difference across PQ= $B v L - \frac{1}{2} B v L = \frac{1}{2} B v L$

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The loop shown moves with a velocity v in a uniform magnetic field of magnitude B, directed into the paper. The potential difference between points P and Q is e. Then

The loop shown moves with a velocity $v$ in a uniform magnetic field of magnitude $B$ , directed into the paper. The potential difference between points $P$ and $Q$ is $e$ . Then