A loop ABCDA, carrying current $I = 12\ \text{A}$, is placed in a plane and consists of two semicircular segments of radius $R_1 = 6\pi\ \text{m}$ and $R_2 = 4\pi\ \text{m}$ joined by the straight segments AB and CD, as shown in the figure. The magnitude of the resultant magnetic field at the centre O is $k\times10^{-7}\ \text{T}$. The value of $k$ is ______. (Given $\mu_0 = 4\pi\times10^{-7}\ \text{T m A}^{-1}$)
Numerical value type. Enter your answer.
Answer: 1
The straight parts AB and CD lie along lines through O, so they give no field at O.
A semicircle of radius $R$ gives $\dfrac{\mu_0I}{4R}$ at its centre. The current goes round the two semicircles in opposite senses, so their fields oppose:
$$B = \frac{\mu_0I}{4}\left(\frac1{R_2} - \frac1{R_1}\right) = \frac{4\pi\times10^{-7}\times12}{4}\left(\frac1{4\pi} - \frac1{6\pi}\right)$$
$$= 12\pi\times10^{-7}\times\frac{1}{12\pi} = 1\times10^{-7}\ \text{T}$$
So $k = 1$.
Solution by Sreeraj P, M.Sc Physics