Electro Magnetic Induction And Alternating Currents Question 446
Question: A conducting square loop is placed in a magnetic field B with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate $ \alpha $ . The induced emf in the loop at an instant when its side is ‘a’ is
Options:
A) $ 2a\alpha B $
B) $ a^{2}\alpha B $
C) $ 2a^{2}\alpha B $
D) $ a\alpha B $
Show Answer
Answer:
Correct Answer: A
Solution:
- At any time t, the side of the square a
$ =(a_{0}-\alpha t) $ ,
where $ a_{0}=side $ at t=0.
At this instant, flux through the square: $ \phi =BA, $
$ \cos 0^{o}=B{{(a_{0}-\alpha t)}^{2}} $
$ \therefore $ emf induced
$ E=-\frac{d\phi }{dt} $
$ \Rightarrow ,E=-B.2(a_{0}-\alpha t)(0-\alpha )=+2\alpha aB $
