Question

A circular coil of radius $5cm$ has $500$ turns of a wire. The approximate value of the coefficient of self-induction of the coil will be

• A. $25mH$
• B. $25\times 10^{-3}mH$
• C. $50\times 10^{-3}mH$
• D. $50\times 10^{-3}H$

Answer 1

Answer: (A)

$\varphi =Li\Rightarrow NBA=Li$
Since magnetic field at the centre of circular coil carrying current is given by $B=\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi Ni}{r}$
$\therefore N\cdot \frac{{\mu }_0}{4\pi }\cdot \frac{2\pi Ni}{r}\cdot \pi r^2=Li$
$\Rightarrow L=\frac{{\mu }_0{N}^2\pi r}{2}$
Hence, self-inductance of a coil
$=\frac{4\pi \times 10^{-7}\times 500\times 500\times \pi \times 0.05}{2}$
$=25mH$