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Q.
A circular coil of radius $5\, cm$ has $500$ turns of a wire. The approximate value of the coefficient of self-induction of the coil will be
Electromagnetic Induction
Solution:
$\phi=L i \Rightarrow N B A=L i$
Since magnetic field at the centre of circular coil carrying current is given by $B=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi N i}{r}$
$\therefore N \cdot \frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi N i}{r} \cdot \pi r^{2}=L i$
$\Rightarrow L=\frac{\mu_{0} N^{2} \pi r}{2}$
Hence, self-inductance of a coil
$=\frac{4 \pi \times 10^{-7} \times 500 \times 500 \times \pi \times 0.05}{2}$
$=25\, mH$