Question

In the figure shown below, the magnetic induction at point $\text{O}$ is,

• A. $\frac{{\mu }_{0}I}{4\pi r}$
• B. $\frac{{\mu }_{0}I}{4r}+\frac{{\mu }_{0}I}{2\pi r}$
• C. $\frac{{\mu }_{0}I}{4r}+\frac{{\mu }_{0}I}{4\pi r}$
• D. $\frac{{\mu }_{0}I}{4r}-\frac{{\mu }_{0}I}{4\pi r}$

Answer 1

The correct option is C $\frac{{\mu }_{0}I}{4r}+\frac{{\mu }_{0}I}{4\pi r}$


The magnetic field due to a straight conductor is,

$B=\frac{{\mu }_{0}I}{4\pi r}[\sin {\theta }_1+\sin {\theta }_2]$

From the diagram,

$B_1=0[\because {\theta }_1={\theta }_2=0^{\circ }]$

$B_3=\frac{{\mu }_{0}I}{4\pi r}[\sin {\theta }_1+\sin {\theta }_2]=\frac{{\mu }_{o}I}{4\pi r}[\because {\theta }_1=0^{\circ };{\theta }_2={90}^{\circ }]⨀$

The magnetic field at point $O$, due to the arc, is,

$B_2=\frac{{\mu }_{o}I}{2r}\left(\frac{\pi }{2\pi }\right)=\frac{{\mu }_{0}I}{4r}[\because \theta =\pi ]⨀$

The net magnetic field at $\text{O}$,

$B_{net}=B_1+B_2+B_3=\frac{{\mu }_{o}I}{4r}+\frac{{\mu }_{o}I}{4\pi r}$

Hence, $\left(C\right)$ is the correct answer.

Why this question ? To understand the concept of magnetic field at a point due to combination of straight conductor and circular arc.