Question

A circular coil has $500$ turns of wire and its radius is $5cm$. The self inductance of the coil is:-

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

Answer 1

The correct option is B $25mH$

$\phi =Li$

$\Rightarrow NBA=Li$ Since magnetic field at the center of circular coil

carring current is given by $B=\frac{{\mu }_0}{4\pi }.\frac{2\pi Ni}{r}$

$\therefore$ $N.\frac{{\mu }_0}{4\pi }.\frac{2\pi Ni}{r}=Li$

$\Rightarrow L=\frac{{\mu }_0{N}^2\pi r}{2}$

Hence,

self conductance of coil is

$=\frac{4\pi \times 10\times 500\times 500\times \pi \times 0.05}{2}$

$=25mH$

Hence,

self conductance of coil is $25mH$

So option $B$ is correct answer.