Electro Magnetic Induction And Alternating Currents Question 445

Question: The magnitude of the earth’s magnetic field at a place is $ B_{0} $ and the angle of dip is $ \delta $ . A horizontal conductor of length l lying along the magnetic north-south moves eastwards with a velocity v. The emf induced across the conductor is [Kerala PET 2005]

Options:

A) Zero

B) $ B_{0}l,v\sin \delta $

C) $ B_{0}l,v $

D) $ B_{0}l,v\cos \delta $

Show Answer

Answer:

Correct Answer: B

Solution:

When a conductor lying along the magnetic north-south, moves eastwards it will cut vertical component of $ B_{0} $ . So induced emf $ e=vB_{V}l $ $ =v(B_{0}\sin \delta ,l) $ $ =B_{0}l,v\sin \delta $ .



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