Question

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

• A. $Zero$
• B. $1\mu V$
• C. $2\mu V$
• D. $10\mu V$

Answer 1

The correct option is D $10\mu V$

Rod is moving towards east, so induced emf across it's end will be $e=B_{V}vl=\left(B_{H}tan\varphi \right)vl$
$\therefore e=3\times {10}^{-4}\times \frac{4}{3}\times (10\times {10}^{-2})\times 0.25={10}^{-5}V=10\mu V$