Question

The current in the wire shown in the figure is $(1/\sqrt{2})A$. The magnetic field at point $P$ is $k\times 10^{-5}T$. The value of $k$ is

Answer 1

Answer: 2

The magnetic field at $P$ due to segments $AB$ and $EF$ are zero,
due to segments $BC$ and $DE$ are $\frac{{\mu }_{0}I}{4\pi r}\left(\sin 45^{\circ }\right)\otimes$
each and due to segment $CD$ is $\frac{{\mu }_{0}I}{4\pi r}\left(\sin 45^{\circ }+\sin 45^{\circ }\right)\otimes$,
where $r=1cm$.

Net magnetic field $=\frac{{\mu }_{0}I}{4\pi r}\left[2\times \sin 45^{\circ }+2\times \sin 45^{\circ }\right]$
$=\frac{10^{-7}\times (1/\sqrt{2})}{10^{-2}}\times 2\sqrt{2}$
$=2\times 10^{-5}T$
$\therefore k=2$