Two concentric coils of 20 turns each are placed in same plane. Their radii are $30 \mathrm{~cm}$ and $60…
- $\frac{8}{3} \mu_0$
- $\frac{2}{3} \mu_0$
- $\frac{5}{3} \mu_0$
- $\frac{10}{3} \mu_0$
Solution

As $\quad B_{\text {centre }}=\frac{\mu_0 n I}{2 r}$ So, Magnetic field at $\mathrm{O}$, $ \begin{aligned} & B_O=\frac{\mu_0}{2}\left[\frac{n_1 I_1}{r_1}-\frac{n_2 I_2}{r_2}\right] \\ & n_1=n_2=20 \\ & B_O=\frac{\mu_0}{4 \pi} \cdot 2 \pi \times 20\left[\frac{0.4}{30 \times 10^{-2}}-\frac{0.6}{60 \times 10^{-2}}\right] \\ & =\frac{\mu_0}{2} \times 20\left[\frac{0.8-0.6}{60 \times 10^{-2}}\right]=\frac{10}{3} \mu_0 \end{aligned} $
Asked in: AP EAMCET 2017 (26 Apr Shift 1)
Practice more Magnetic Fields due to Electric Current questions on Aicharya