Electro Magnetic Induction And Alternating Currents Question 428
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 [MP PMT 1992; AIEEE 2004]
Options:
A) $ 5\ mV $
B) $ 5\times {{10}^{-4}}V $
C) $ 50\ mV $
D) $ 50\ \mu V $
Show Answer
Answer:
Correct Answer: D
Solution:
$ 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 $
BETA
