Question

A wire carrying current $I$ has the shape as shown in adjoining figure.
Linear parts of the wire are very long and parallel to . $X$-axis while semicircular portion of radius $R$ is lying in $Y-Z$ plane. Magnetic field at point $O$ is

• A. $\vec{B}=-\frac{{\mu }_{0}I}{4\pi }\frac{I}{R}(\pi \hat{i}+2\hat{k})$
• B. $\vec{B}=\frac{{\mu }_0}{4\pi }\frac{I}{R}(\pi \hat{i}-2\hat{k})$
• C. $\vec{B}=\frac{{\mu }_0}{4\pi }\frac{I}{R}(\pi \hat{i}+2\hat{k})$
• D. $\vec{B}=-\frac{{\mu }_0}{4\pi }\frac{I}{R}(\pi \hat{i}-2\hat{k})$

Answer 1

Answer: (A)

Given situation is shown in the figure.

Parallel wires $1$ and $3$ are semi-infinite, so magnetic field at $O$ due to them
$\overrightarrow{B_1}=\overrightarrow{B_3}=-\frac{{\mu }_{0}I}{4\pi R}\hat{k}$
Magnetic field at $O$ due to semi-circular are in
YZ-plane is given by $\overrightarrow{B_2}=-\frac{{\mu }_{0}I}{4R}\hat{i}$
Net magnetic field at point $O$ is given by
$\vec{B}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}$
$=-\frac{{\mu }_{0}I}{4\pi R}\hat{k}-\frac{{\mu }_{0}I}{4R}\hat{i}-\frac{{\mu }_{0}I}{4\pi R}\hat{k}$
$=-\frac{{\mu }_{0}I}{4\pi R}(\pi \hat{i}+2\hat{k})$