Question

A wire carrying current $I$ has the shape as shown in 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. $\overrightarrow{B}=-\frac{{\mu }_0}{4\pi }\frac{I}{R}(\pi \hat{i}-2\hat{k})$
• 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 B $\overrightarrow{B}=-\frac{{\mu }_0}{4\pi }\frac{I}{R}(\pi \hat{i}+2\hat{k})$


Magnetic field due to segment '1'
$\overrightarrow{B_1}=\frac{{\mu }_{0}I}{4\pi R}[\sin {90}^{\circ }+\sin \theta ](-\hat{k})$
$\overrightarrow{B_1}=\frac{{\mu }_{0}I}{4\pi R}(-\hat{k})=\overrightarrow{B_3}$
Magnetic field due to segement '2'
$\overrightarrow{B_2}=\frac{{\mu }_{0}I}{4R}(-\hat{i})$
So, magnetic field at centre $\overrightarrow{B_C}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}$
$\Rightarrow \overrightarrow{B_C}=\frac{-{\mu }_{0}I}{4R}(\hat{i}+\frac{2\hat{k}}{\pi })=\frac{-{\mu }_{0}I}{4\pi R}(\pi \hat{i}+2\hat{k})$