Q. A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 rad${\ s}^{- 1}$. If the horizontal component of earth’s magnetic field is $0.2 \times 10^{- 4}\ T,$ then the emf developed between the two ends of the conductor is
A) 5 $\mu V$
B) 50$\ \mu V$
C) 5 mV
D) 50 mV
Explanation
(b) The emf induced between ends of conductor
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = \frac{1}{2}B\omega L^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2} \times 0.2 \times 10^{- 4} \times 5 \times (1)^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.5 \times 10^{- 4}V$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5 \times 10^{- 5}V = 50\mu V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.