Question

A current $I$ flows around a closed path in the horizontal plane as shown in the figure. The path consists of eight arcs with alternating radii $r$ and $2r$. Each segment of arc subtends equal angle at the common center. The magnetic induction at common centre is

• A. $\frac{3}{8}\frac{{\mu }_{0}I}{r}⨂$ ;perpendicular to the plane of the paper and directed inward
• B. $\frac{3}{8}\frac{{\mu }_{0}I}{r}⨀$ perpendicular to the plane of the paper and directed outward
• C. $\frac{1}{8}\frac{{\mu }_{0}I}{r}⨂$ ;perpendicular to the plane of the paper and directed inward.
• D. $\frac{1}{8}\frac{{\mu }_{0}I}{r}⨀$;perpendicular to the plane of the paper and directed outward.

Answer 1

The correct option is B $\frac{3}{8}\frac{{\mu }_{0}I}{r}⨀$ perpendicular to the plane of the paper and directed outward

Magnetic field due a circular arc at its centre is,

$B=\frac{{\mu }_{0}i}{2r}\left(\frac{\theta }{2\pi }\right)....\left(1\right)$

As we can see, the complete circle subtends an $2\pi$ at its centre, but here four arcs of radii $2r$ and other four arcs of radii $r$ since each arc subtends equal angle at the centre. Thus, four arcs of radii $2r$ toghether subtends an angle $\pi$ at the center and other four arcs of radii $r$ toghether subtends an angle $\pi$ at the center.

Using (1), field due to the arcs of radius $2r$ is,

$B_{2r}=\frac{{\mu }_{0}i}{2\left(2r\right)}\left(\frac{\theta }{2\pi }\right)=\frac{{\mu }_{0}i\pi }{8\pi r}=\frac{{\mu }_{0}i}{8r}\odot$

Field due to the arcs of radius $r$ is,

$B_r=\frac{{\mu }_{0}i}{2r}\left(\frac{\theta }{2\pi }\right)=\frac{{\mu }_{0}i\pi }{4\pi r}=\frac{{\mu }_{0}i}{4r}\odot$

Net magnetic field at the center is,

$B_{net}=\frac{{\mu }_{0}i}{4r}+\frac{{\mu }_{0}i}{8r}=\frac{3{\mu }_{0}i}{8r}\odot$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.