Question

In the given figure magnetic field at the center of ring $O$ is $\frac{8}{\sqrt{2}}T$ . Now it is turned through $90^{\circ }$ about $x{x}^{\prime }$ axis, so that two semicircular parts are mutually perpendicular. The new value of magnetic field (in Tesla) at centre is $\sqrt{n}$ . Find $n$ .

Answer 1

Answer: 04.00

Given, $\frac{{\mu }_{0}i}{2R}=\frac{8}{\sqrt{2}}T$
In $2^{\text{nd }}$ case,
$|\vec{B}|=\sqrt{{\left(\frac{{\mu }_{0}i}{4R}\right)}^2+{\left(\frac{{\mu }_{0}i}{4R}\right)}^2}=\frac{{\mu }_{0}i}{2\sqrt{2}R}=4\mathrm{T}$