Electro Magnetic Induction And Alternating Currents Question 485
Question: A superconducting loop of radius R has self-inductance L. A uniform and constant magnetic field B is applied perpendicular to the plane of the loop. Initially current in this loop is zero. The loop is rotated by $ 180{}^\circ $ . The current in the loop after rotation is equal to
Options:
A) zero
B) $ \frac{B\pi R^{2}}{L} $
C) $ \frac{2B\pi R^{2}}{L} $
D) $ \frac{B\pi R^{2}}{2L} $
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Answer:
Correct Answer: C
Solution:
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Flux can’t change in a superconducting loop.
$ \Delta \phi =2\pi R^{2}.B $
Initially current was zero, so self-flux was zero.
$ \therefore $ Finally $ Li=2\pi R^{2}\times B $ $ i=\frac{2\pi R^{2}\times B}{L} $
