A) 2 L
B) L
C) 4 L
D) 8 L
Correct Answer: A
Solution :
\[L=\frac{{{\mu }_{0}}{{N}^{2}}\pi {{r}^{2}}}{\ell }\] |
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. |
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