Question

A straight wire is bent in a shape shown in the diagram. It carries a steady current $I$. The magnetic field at the Center $O$ of the arcs will be,

• A. $\frac{{\mu }_{o}I}{4}\left(\frac{r_1}{r_2^2}\right)\otimes$
• B. $\frac{{\mu }_{o}I}{4}\left(\frac{1}{r_2}\right)\odot$
• C. $\frac{{\mu }_{o}I}{4}\left(\frac{r_1+r_2}{r_1{r}_2}\right)\otimes$
• D. $\frac{{\mu }_{o}I}{4}\left(\frac{r_1+r_2}{r_1{r}_2}\right)\odot$

Answer 1

The correct option is C $\frac{{\mu }_{o}I}{4}\left(\frac{r_1+r_2}{r_1{r}_2}\right)\otimes$

Magnetic field due to the arc of radius $r_1$ is,

$B_1=\frac{{\mu }_{o}I}{4r_1}\text{Inward}(\otimes )$

Magnetic field due to the arc of radius $r_2$ is,

$B_2=\frac{{\mu }_{o}I}{4r_2}\text{Inward}(\otimes )$

The magnetic field at point $O$ due to straight segments will be equal to zero.

The net magnetic field at point $O$ due to the entire conductor will be,

$B_{net}=B_1+B_2=\frac{{\mu }_{o}I}{4}\left(\frac{r_1+r_2}{r_1{r}_2}\right)\text{Inward}(\otimes )$

Hence, option (c) is the correct answer.