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
$=\frac{{\mu }_0{i}_1}{2R}(\frac{{\theta }_1}{2\pi })-\frac{{\mu }_0{i}_2}{2R}(\frac{{\theta }_2}{2\pi })$
$=\frac{{\mu }_0}{2R}(\frac{3\pi }{2\times 2\pi })-\frac{{\mu }_0{i}_2}{2R}(\frac{\pi }{2\times 2\pi })$
$=\frac{{\mu }_0}{2R}[\frac{3i_1}{4}-\frac{i_2}{4}]$
$=\frac{{\mu }_0}{2R}[\frac{3i_1}{4}-\frac{3i_1}{4}]=0$