The magnetic field at the centre of a wire loop formed by two semicircular wires of radii $R_1 = 2\pi\ \text{m}$ and $R_2 = 4\pi\ \text{m}$ carrying current $I = 4\ \text{A}$ as per figure given below is $\alpha\times10^{-7}\ \text{T}$. The value of $\alpha$ is ______. (Centre O is common for all segments)
Numerical value type. Enter your answer.
Answer: 3
The straight segments lie along lines through O, so they give no field at O. The current goes round the loop in one sense, so the two semicircles give fields in the same direction:
$$B = \frac{\mu_0 I}{4R_1} + \frac{\mu_0 I}{4R_2} = \frac{4\pi\times10^{-7}\times4}{4}\left(\frac{1}{2\pi} + \frac{1}{4\pi}\right)$$
$$B = 4\pi\times10^{-7}\times\frac{3}{4\pi} = 3\times10^{-7}\ \text{T} \;\Rightarrow\; \alpha = 3$$
Solution by Sreeraj P, M.Sc Physics