Question

A metal conductor of length 1m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is $0.2 \times 10^{- 4}T$, then the e.m.f. developed between the two ends of the conductor is

• A. $5\mspace{6mu} mV$
• B. $5 \times 10^{- 4}V$
• C. $50\mspace{6mu} mV$
• D. $50\mspace{6mu}\mu V$

Answer 1

Answer: (D)

$50\mspace{6mu}\mu V$

Answer 2

$e = \frac{1}{2}B\omega r^{2} = \frac{1}{2} \times 0.2 \times 10^{- 4} \times 5 \times (1)^{2} = 50\mu V$