A) zero
B) \[\frac{B\pi {{R}^{2}}}{L}\]
C) \[\frac{2B\pi {{R}^{2}}}{L}\]
D) \[\frac{B\pi {{R}^{2}}}{2L}\]
Correct Answer: C
Solution :
[c] 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 \[Li=2\pi {{R}^{2}}\times B\] \[i=\frac{2\pi {{R}^{2}}\times B}{L}\]You need to login to perform this action.
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