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 $2$
$B_{2}=\frac{\mu_{0} I}{4 R}(-\hat{i})=\frac{-\mu_{0} I}{4 \pi R}(\pi \hat{i})$
$\therefore \overrightarrow{ B }$ at centre
$\overrightarrow{ B }_{ c }=\overrightarrow{ B _{1}}+\overrightarrow{ B _{2}}+\overrightarrow{ B _{3}}=\frac{-\mu_{0} I }{4 \pi R }(\pi \hat{ i }+2 \hat{ k })$.