A current of $5\ \text{A}$ exists in a square loop of side $\dfrac{1}{\sqrt2}\ \text{m}$. Then the magnitude of the magnetic field $B$ at the centre of the square loop will be $p\times10^{-6}\ \text{T}$, where the value of $p$ is ______. [Take $\mu_0 = 4\pi\times10^{-7}\ \text{T m A}^{-1}$]
Numerical value type. Enter your answer.
Answer: 8
Each side is at distance $d = a/2$ from the centre and subtends $45^\circ$ on each side:
$$B_{side} = \frac{\mu_0I}{4\pi d}(\sin45^\circ + \sin45^\circ) = \frac{\mu_0I\sqrt2}{4\pi(a/2)}$$
Four sides:
$$B = 4\cdot\frac{\sqrt2\,\mu_0I}{2\pi a} = \frac{2\sqrt2\,\mu_0I}{\pi a} = \frac{2\sqrt2\times4\pi\times10^{-7}\times5}{\pi\times\frac{1}{\sqrt2}} = 2\sqrt2\cdot\sqrt2\times4\times10^{-7}\times5 = 8\times10^{-6}\ \text{T}$$
$p = 8$
Solution by Sreeraj P, M.Sc Physics