Question

The horizontal component of earth's magnetic field at a place is $3\times {10}^{-4}T$ and the dip is ${\tan }^{-1}\left(\frac{4}{3}\right)$. A metal rod of length $0.25m$ is placed in $N-S$ direction and is moved at a constant speed of $10cm{s}^{-1}$ towards the east. The emf induced in the rod will be

• A. $1\mu V$
• B. $5\mu V$
• C. $7\mu V$
• D. $10\mu V$

Answer 1

Answer: (D)

$10\mu V$

Rod is moving towards east, so induced emf across its end will be
$e=B_{v}vl=\left(B_H\tan \theta \right)vl[\therefore \tan \theta =\frac{B_V}{B_H}]$
$=3\times {10}^{-4}\times \frac{4}{3}\times 10\times {10}^{-3}\times 0.25$
$={10}^{-5}V=10\mu V$.