Q. The horizontal component of the 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 $0.25\ m$ 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) $5\ \mu V$
D) $10\ \mu V$
Explanation
(d) Rod is moving towards east, so induced emf across it’s end will be $e = B_{V}vl = \left( B_{H}\tan\phi \right)vl$
$\therefore e = 3 \times 10^{- 4} \times \frac{4}{3} \times \left( 10 \times 10^{- 2} \right) \times 0.25 = 10^{- 5}V = 10\mu V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.