A) 25 millihenry
B) \[25\times {{10}^{-3}}\]millihenry
C) \[50\times {{10}^{-3}}\]millihenry
D) \[50\times {{10}^{-3}}\]henry
Correct Answer: A
Solution :
\[\varphi =Li\Rightarrow NBA=Li\] 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 \[=\frac{4\pi \times {{10}^{-7}}\times 500\times 500\times \pi \times 0.05}{2}=25\ mH\]
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