Question

The figure shows a square loop $PQRS$ with an edge length $a$ . The resistance of the wire $PQR$ is $r$ and that of $PSR$ is $2r$ . The value of the magnetic field at the centre of the loop is found to be $\frac{\sqrt{2}{\mu }_{0}i}{k\pi a}$ , then what is the value of $k$ ?

Answer 1

Answer: 3

The magnetic field due to a straight current-carrying wire is
$B=\frac{{\mu }_{0}I}{4\pi d}\left[sin\alpha +sin\beta \right]$
$B_{\mathrm{PQR}}=2\times \frac{{\mu }_0\left(\frac{2i}{3}\right)}{4\pi \left(\frac{a}{2}\right)}\left[\sin 45^{\circ }+\sin 45^{\circ }\right]$
$B_{\mathrm{PSR}}=2\times \frac{{\mu }_0\left(\frac{i}{3}\right)}{4\pi \left(\frac{a}{2}\right)}\left[\sin 45^{\circ }+\sin 45^{\circ }\right]$
$B_{net}=B_{PQR}-B_{PSR}$
$\Rightarrow B_{net}=\frac{\sqrt{2}{\mu }_{0}i}{3\pi a}$
$\Rightarrow k=3$