Electro Magnetic Induction And Alternating Currents Question 424

Question: A conductor of 3 m in length is moving perpendicularly to magnetic field of $ {{10}^{-3}} $ tesla with the speed of $ 10^{2}m/s $ , then the e.m.f. produced across the ends of conductor will be [MP PET 1990]

Options:

A) 0.03 volt

B) 0.3 volt

C) $ 3\times {{10}^{-3}}volt $

D) 3 volt

Show Answer

Answer:

Correct Answer: B

Solution:

$ e=Bvl=3\times {{10}^{-3}}\times 10^{2}=0.3\ volt $



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