Question

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

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

$\text{Given:-}$ The situation is represented in the figure

$\text{To find:-}$ Magnetic field at point O

$\text{Solution:-}$ Parallel wires 1 and 3 are semi-infinite, so magnetic field at O due to them

$\overrightarrow{B_1}=\overrightarrow{B_3}=-\frac{{\mu }_{0}I}{4\pi R}\hat{k}$

Now, Magnetic field at O due to semi-circular are in YZ-plane ,

$\overrightarrow{B_2}=-\frac{{\mu }_{0}I}{4R}\hat{i}$

Therefore ,Net magnetic field at point O is given by

$\overrightarrow{B}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}$

= -$\frac{{\mu }_{0}I}{4\pi R}\hat{k}-\frac{{\mu }_{0}I}{4R}\hat{i}-\frac{{\mu }_{0}I}{4\pi R}\hat{k}$

= $-\frac{{\mu }_{0}I}{4\pi R}(\pi \hat{i}+2\hat{k})$

$\text{Hence the correct option is D}$