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 $\mathrm{AC}$, $
B_{A C}=\frac{\mu_0 I}{4 \pi(2 R \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}}$