Question

A straight conductor carrying current $i$ splits into two parts as shown in the figure. The radius of the circular loop is $R$. The total magnetic field at the centre $P$ at the loop is :

• A. Zero
• B. 3$\mu_o$i/32 R, outward
• C. 3$\mu_o$i/32R, inward
• D. $\frac{\mu_{0}i}{2R}$,inward

Answer 1

Answer: (A)

Magnetic field due to $i_1 = \frac{\mu_0 i_1}{2R} \frac{\theta_1}{2\pi}$
(Into the plane)
Magnetic field due to $i_2 = \frac{\mu_0 i_2}{2R} \frac{\theta_2}{2\pi}$
(out of the plane )
For parallel combination $\frac{i_1}{i_2} = \frac{\rho l_2}{A} \times \frac{A}{\rho l_1} = \frac{l_1}{l_2}$
$ \Rightarrow \frac{i_1}{i_2} = \frac{\frac{1}{4} (2\pi R)}{\frac{3}{4} (2\pi R)}$
$ = \frac{1}{3} $
$\Rightarrow i_1 = \frac{i_2}{3}$
$\Rightarrow i_2 = 3i_1$
$\therefore $ Net magnetic field
$ = \frac{\mu_0i_1}{2R}(\frac{\theta_1}{2\pi}) - \frac{\mu_0 i_2}{2R}(\frac{\theta_2}{2\pi})$
$ = \frac{\mu_0}{2R}(\frac{3\pi}{2\times 2\pi}) - \frac{\mu_0i_2}{2R}(\frac{\pi}{2\times 2\pi})$
$ = \frac{\mu_0}{2R}[\frac{3i_1}{4} - \frac{i_2}{4}]$
$ = \frac{\mu_0}{2R}[\frac{3i_1}{4} - \frac{3i_1}{4}]= 0$