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

Answer 1

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


Magnetic field due to segment $1$
$\overrightarrow{B}={\overrightarrow{B}}_1=\frac{{\mu }_{0}I}{4\pi R}[sin{90}^{\circ }+sin{0}^{\circ }](-\hat{k})$
${\overrightarrow{B}}_1=\frac{-{\mu }_{0}I}{4\pi R}\left(\hat{k}\right)$
Magnetic field due to segment $2$
$B_2=\frac{{\mu }_{o}I}{4\pi R}\pi (-\hat{i})=\frac{-{\mu }_{o}I}{4\pi R}\left(\pi \hat{i}\right)$
Magnetic field due to segment $3$
$\overrightarrow{B}={\overrightarrow{B}}_3=\frac{{\mu }_{0}I}{4\pi R}[sin{90}^{\circ }+sin{0}^{\circ }](-\hat{k})$
${\overrightarrow{B}}_3=\frac{-{\mu }_{0}I}{4\pi R}\left(\hat{k}\right)$
Net magnetic field at O
${\overrightarrow{B}}_c={\overrightarrow{B}}_1+{\overrightarrow{B}}_2+{\overrightarrow{B}}_3$
$=\frac{-{\mu }_{o}I}{4\pi R}(\pi \hat{i}+2\hat{k})$