Question

A current $I$ flows around a closed path in the horizontal plane of the circle 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 centre $P$. The magnetic field produced by current path at point $P$ 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 A $\frac{3}{8}\frac{{\mu }_{0}I}{r}$ ; perpendicular to the plane of the paper and directed inward.

Magnetic field due to current-carrying wire is inside

Magnetic field due to any arc $\frac{\mu i\theta }{2r2\pi }$

Magnetic field due to 2r radius=$\frac{\mu i\pi }{2\cdot 2r2\pi }=\frac{\mu i\pi }{8r\pi }$

Magnetic field due to r radius=$\frac{\mu i\pi }{2r2\pi }=\frac{\mu i\pi }{4r\pi }$

Net Magnetic field=$\frac{\mu i\pi }{8r\pi }+\frac{\mu i\pi }{4r\pi }$

=$\frac{3ui}{8r}$

Hence, Option A is answer