Question 88 of 153

A wire of length $1m$ moving with velocity $8m{s}^{-1}$ at right angles to a magnetic field of $2T$. The magnitude of induced emf, between the ends of wire will be ___________.

A$20V$B$8V$C$12V$D$16V$

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Answer: (D) $16V$

The expression of motional emf is $e=BLv\sin \theta$, where, $\theta$ is angle between magnetic field and direction of motion.
Here, $\theta =90^{\circ}$,
So, induced emf across the ends of wire is $e=BLv\sin 90^{\circ} =BLv$
$=2\times 1\times 8 =16V$

JEE Main Physics practice question on Electromagnetic Induction (Motional and Rotational EMF), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.