A current loop consists of two identical semicircular parts each of radius $R$, one lying in the $x-y$ plane…

A current loop consists of two identical semicircular parts each of radius $R$, one lying in the $x-y$ plane and the other in $x-z$ plane. If the current in the loop is $i$. The resultant magnetic field due to the two semicircular parts at their common centre is
  1. $\frac{\mu_0 i}{2 \sqrt{2} R}$
  2. $\frac{\mu_0 \mathrm{i}}{2 R}$
  3. $\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}$
  4. $\frac{\mu_0 \mathrm{i}}{\sqrt{2} \mathrm{R}}$

Solution

The magnetic field at centre $O$ for each semicircular parts each of radius $R$, $\mathrm{B}_1=\mathrm{B}_2=\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}$ The magnetic field at their common centre $\begin{aligned} \overrightarrow{\mathrm{B}} & =\overrightarrow{\mathrm{B}}_1+\overrightarrow{\mathrm{B}}_2 \\ \mathrm{~B} & =\sqrt{\mathrm{B}_1^2+\mathrm{B}_2^2} \\ & =\sqrt{\left(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}\right)^2+\left(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}\right)^2} \\ \mathrm{~B} & =\frac{\mu_0 \mathrm{i}}{2 \sqrt{2} \mathrm{R}} \end{aligned}$ :

Asked in: NEET 2010 (Mains)

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