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.
You will be redirected in
3 sec