Question

A hairpin-like shape as shown in the figure is made by bending a long current-carrying wire. What is the magnitude of a magnetic field at point $P$ which lies on the center of the semicircle?

• A. $\begin{array}{l}\frac{{\mu }_{0}I}{4\pi r}(2-\pi )\end{array}$
• B. $\begin{array}{l}\frac{{\mu }_{0}I}{4\pi r}(2+\pi )\end{array}$
• C. $\begin{array}{l}\frac{{\mu }_{0}I}{2\pi r}(2+\pi )\end{array}$
• D. $\begin{array}{l}\frac{{\mu }_{0}I}{2\pi r}(2-\pi )\end{array}$

Answer 1

The correct option is B

$\begin{array}{l}\frac{{\mu }_{0}I}{4\pi r}(2+\pi )\end{array}$

Step 1. Given data

A hairpin is made by bending a long current-carrying wire.

Step 2. Finding the magnitude of the magnetic field at the point $P$:

Magnetic field due to each of the straight wire $=\begin{array}{l}\frac{{\mu }_{0}I}{4\pi r}\end{array}$

Magnetic field due to semicircular arc $=\begin{array}{l}\frac{{\mu }_{0}I}{4r}\end{array}$

where, $I$ is current through the wire

${\mu }_0$ permeability of free space

$r$ distance from the conductor

Thus, the total magnetic field at a point $P$ is given by

$\begin{array}{lcl}& & \begin{array}{l}B=\frac{{\mu }_{0}I}{4\pi r}+\frac{{\mu }_{0}I}{4\pi r}+\frac{{\mu }_{0}I}{4r}\end{array}\\ & & \begin{array}{l}B=\frac{{\mu }_{0}I}{2\pi r}+\frac{{\mu }_{0}I}{4r}\end{array}\\ & & \begin{array}{l}B=\frac{{\mu }_{0}I}{4\pi r}(2+\pi )\end{array}\end{array}$

Hence, option$\left(B\right)$ is correct.