A) \[25\text{ }mH\]
B) \[25\times {{10}^{-3}}\,mH\]
C) \[50\times {{10}^{-3}}\,mH\]
D) \[50\times {{10}^{-3}}\,H\]
Correct Answer: A
Solution :
[a] \[\phi =Li\Rightarrow NBA=Li\] Since magnetic field at the centre of circular coil carrying current is given by \[B=\frac{{{\mu }_{0}}}{4\pi }\], \[\frac{2\pi Ni}{r}\] \[\therefore \,\,\,\,N.\frac{{{\mu }_{0}}}{4\pi }\,\,.\frac{2\pi Ni}{r}.\,\,\pi {{r}^{2}}=Li\Rightarrow L=\frac{{{\mu }_{0}}{{N}^{2}}\pi r}{2}\] Hence self-inductance of a coil \[\frac{4\pi \times {{10}^{-7}}\times 500\times 500\times \pi \times 0.05}{2}=25mH\]
You need to login to perform this action.
You will be redirected in
3 sec
