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

Answer 1

Answer: (B)

Magnetic field due to segment ' $1$'
${\overrightarrow{B}}_1=\frac{{\mu }_{0}I}{4\pi R}\left[\sin 90^{\circ }+\sin 0^{\circ }\right](-\hat{k})$
$=\frac{-{\mu }_{0}I}{4\pi R}(k)={\overrightarrow{B}}_3$
Magnetic field due to segment $2$
$B_2=\frac{{\mu }_{0}I}{4R}(-\hat{i})=\frac{-{\mu }_{0}I}{4\pi R}(\pi \hat{i})$
$\therefore \overrightarrow{B}$ at centre
${\overrightarrow{B}}_c=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}=\frac{-{\mu }_{0}I}{4\pi R}(\pi \hat{i}+2\hat{k})$.