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

Answer 1

The correct option is B $\overrightarrow{B}=-\frac{{\mu }_0}{4\pi }\frac{I}{R}(\pi \hat{i}+2\hat{k})$

Magnetic field due to segment ${}^{\prime }{1}^{\prime }$
$\overrightarrow{B_1}=\frac{{\mu }_{0}I}{4\pi R}[\sin {90}^{\circ }+\sin 0^{\circ }](-\hat{k})$
$=\frac{-{\mu }_{0}I}{4\pi R}\left(\hat{k}\right)=\overrightarrow{B_3}$
Magnetic field due to segment $2$
$B_2=\frac{{\mu }_{0}I}{4R}(-\hat{i})=\frac{-{\mu }_{0}I}{4\pi R}\left(\pi \hat{i}\right)$
$\therefore \overrightarrow{B}$ at centre
$\overrightarrow{B_c}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}=\frac{-{\mu }_{0}I}{4\pi R}(\pi \hat{i}+2\hat{k})$.