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 $ .
