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 $



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