Question

A conducting wire carrying current is arranged as shown. The magnetic field at $'O'$

• A. $\frac{\mu_0 i}{12} \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$
• B. $\frac{\mu_0 i}{12} \left[ \frac{1}{R_1} + \frac{1}{R_2} \right]$
• C. $\frac{\mu_0 i}{6} \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$
• D. $\frac{\mu_0 i}{6} \left[ \frac{1}{R_1} + \frac{1}{R_2} \right]$

Answer 1

Answer: (A)

$\frac{\mu_0 i}{12} \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$

Answer 2

Key Idea The magnetic field at the centre due to an arc carrying-current is given by the formula $B=\frac{\mu_{0}}{4 \pi}\left(\frac{i}{R}\right) \theta$

Here, $\theta=60^{\circ}=\frac{\pi}{3}$
The magnetic field due to $R_{1}$ radius part.
$B_{1}=\frac{\mu_{0}}{4 \pi}\left(\frac{i}{R_{1}}\right) \frac{\pi}{3}$
Similarly due $R_{2}$,
$B_{2}=\frac{\mu_{0}}{4 \pi}\left(\frac{i}{R_{2}}\right) \frac{\pi}{3}$
Net magnetic field,
$= B = B _{ 1 }- B _{2}$
$=\frac{\mu_{0}}{4 \pi}\left(\frac{i}{R_{1}}\right) \frac{\pi}{3}-\frac{\mu_{0}}{4 \pi}\left(\frac{i}{R_{2}}\right) \frac{\pi}{3}$
$=\frac{\mu_{0}}{4 \pi} i \frac{\pi}{3}\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
$=\frac{\mu_{0} i}{12}\left(\frac{1}{ R _{1}}-\frac{1}{ R _{ z }}\right)$