Electro Magnetic Induction And Alternating Currents Question 464
Question: The magnetic field in a region is given by $ B=B_{0}( 1+\frac{x}{a} )\hat{k} $ . A square loop of edge-length d is placed with its edges along the x and y-axes. The loop is moved with a constant velocity $ v=v_{0}\hat{i}. $ The emf induced in the loop is:
Options:
A) zero
B) $ v_{0}B_{0}d $
C) $ \frac{v_{0}B_{0}d^{3}}{a^{2}} $
D) $ \frac{v_{0}B_{0}d^{2}}{a} $
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Answer:
Correct Answer: D
Solution:
-
At $ x=0 $ , $ B=B_{0} $ , so $ e_{1}=B_{0}v_{0}d $
At, $ x=d $ , $ B=B_{0}( 1+\frac{d}{a} ) $ , so $ e_{2}=B_{0}( 1+\frac{d}{a} )v_{0}d $
Now $ e_{net}=e_{2}-e_{1}=\frac{B_{0}v_{0}d^{2}}{a} $
BETA
