A) 2L
B) L
C) 4L
D) 8L
Correct Answer: A
Solution :
[a] \[L=\frac{{{\mu }_{0}}{{N}^{2}}\pi {{r}^{2}}}{\ell }\] Length of wire=N \[2\pi r\]=constant (=C, suppose) \[\therefore \,\,\,\,L={{\mu }_{0}}{{\left( \frac{C}{2\pi r} \right)}^{2}}\frac{\pi {{r}^{2}}}{\ell }\] \[\therefore \,\,\,\,L\propto \frac{1}{\ell }\] \[\therefore \] Self-inductance will become 2L.You need to login to perform this action.
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