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

Answer: (D)

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

$\text{Solution:-}$

Magnetic field due to staright wire 1 is given by,

${\overrightarrow{B}}_1=\frac{{\mu }_{0}I}{4\pi R}[sin{90}^o+sin{0}^o]\left(-\overrightarrow{k}\right)$

=$\frac{-{\mu }_{0}I}{4\pi R}\hat{k}={\overrightarrow{B}}_3$

Magnetic field due to semicircular wire 2 is given by,

$B_2=\frac{-{\mu }_{0}I}{4R}(-\hat{i})=\frac{-{\mu }_{0}I}{4\pi R}\left(\pi \hat{i}\right)$

Also,Magnetic field $\overrightarrow{B}$ at centre,

${\overrightarrow{B}}_c={\overrightarrow{B}}_1+{\overrightarrow{B}}_2+{\overrightarrow{B}}_3$

$⟹{\overrightarrow{B}}_c=\frac{-{\mu }_{0}I}{4\pi R}\left(\pi \hat{i}+2\hat{k}\right)$

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