Question

A current-carrying circular coil of 100 turns and radius 5.0 cm produces a magnetic field of 6.0 × 10⁻⁵ T at its centre. Find the value of the current.

Answer 1

Given:
No. of turns, n = 100
Radius of the loop, r = 5 cm = 0.05 m
Magnetic field intensity, B = 6.0 × 10⁻⁵ T
Now, let the magnitude of current be i.
$$\begin{gathered} \mathrm{Using}B=\frac{{\mathrm{\mu }}_0\mathrm{ni}}{2\mathrm{r}},\mathrm{we}\mathrm{get} \\ 6.0\times {10}^{-7}=\frac{4\mathrm{\pi }\times {10}^{-7}\times {10}^2\times i}{2\times 0.05} \\ \Rightarrow i=\frac{60\times {10}^{-7}}{4\mathrm{\pi }\times {10}^{-7}\times {10}^2} \\ =4.777\times {10}^{-2}\approx 48\mathrm{mA} \end{gathered}$$