Question

A current-carrying wire is placed in the grooves of an insulating semicircular disc of radius $R$ as shown. The current enters at point $A$ and leaves from point $B$ . Determine the magnetic field at point $D$ .

• A. $\frac{{\mu }_0\text{I}}{8\pi \text{R}\sqrt{3}}$
• B. $\frac{{\mu }_0\text{I}}{4\pi \text{R}\sqrt{3}}$
• C. $\frac{\sqrt{3}{\mu }_0\text{I}}{4\pi \text{R}}$
• D. none of these

Answer 1

Answer: (B)

Magnetic field due to wire $\mathrm{AC}$, $<br/>B_{AC}=\frac{{\mu }_{0}I}{4\pi (2R\sin 30)}(\cos 30-\cos 90)=\frac{\sqrt{3}{\mu }_{0}I}{8\pi R}$
Magnetic field due to wire BC,
${\mathrm{B}}_{\mathrm{BC}}=\frac{{\mu }_0\mathrm{I}}{4\pi (2\mathrm{R}\sin 60)}(\cos 60-\cos 90)=\frac{{\mu }_0\mathrm{I}}{8\sqrt{3}\pi \mathrm{R}}=\frac{\sqrt{3}{\mu }_0\mathrm{I}}{24\pi \mathrm{R}}$
${\overrightarrow{\mathrm{B}}}_{\text{net }}={\mathrm{B}}_{\mathrm{A}}-{\mathrm{B}}_{\mathrm{B}}$
$=\frac{\sqrt{3}{\mu }_0\mathrm{I}}{8\pi \mathrm{R}}\left(1-\frac{1}{3}\right)$
$=\frac{2\sqrt{3}{\mu }_0\mathrm{I}}{24\pi \mathrm{R}}=\frac{{\mu }_0\mathrm{I}}{4\sqrt{3}\pi \mathrm{R}}$