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. $\frac{{\mu }_{0}I}{8\pi R\sqrt{3}}$
• B. $\frac{{\mu }_{0}I}{4\pi R\sqrt{3}}$
• C. $\frac{\sqrt{3}{\mu }_{0}I}{4\pi R}$
  
• D. None of these

Answer 1

The correct option is B $\frac{{\mu }_{0}I}{4\pi R\sqrt{3}}$

Magnetic field due to $AC$ at $D$ is
$B_{AC}=\frac{{\mu }_{0}I[\sin 60+\sin (0\left)\right]}{4\pi \left(2R\sin 30\right)}=\frac{{\mu }_{0}I\frac{\sqrt{3}}{2}}{4\pi R}=\frac{{\mu }_{0}I\sqrt{3}}{8\pi R}\odot$

Magnetic field due to $CB$ at $D$ is
$B_{CB}=\frac{{\mu }_{0}I(\sin 30+\sin 0)}{4\pi \left(2R\sin 60\right)}$
$B_{CB}=\frac{{\mu }_{0}I(\frac{1}{2}+0)}{4\pi R\sqrt{3}}=\frac{{\mu }_{0}I}{8\pi R\sqrt{3}}\otimes$

Net Magnetic field at $D$ is
But $B_{net}=B_{AC}-B_{CB}=\frac{{\mu }_{0}I}{4\pi R}[\frac{\sqrt{3}}{2}-\frac{1}{2\sqrt{3}}]$
$B_{net}=\frac{{\mu }_{0}I}{4\pi R}$$\frac{(3-1)}{2\sqrt{3}}$
$B_{net}=\frac{{\mu }_{0}I}{4\sqrt{3}\pi R}$
Ans: $B$