Question

Shown in the figure, is a conductor carrying current $I$. The net magnetic induction at the point $\text{O}$ is -

[Point $\text{O}$ is the common center of all the three arcs.]

• A. Zero
• B. $\frac{{\mu }_{0}I\theta }{24\pi R}$
• C. $\frac{11{\mu }_{0}I\theta }{24\pi R}$
• D. $\frac{5{\mu }_{0}I\theta }{24\pi R}$

Answer 1

The correct option is D $\frac{5{\mu }_{0}I\theta }{24\pi R}$


From the figure, we can see that magnetic fields at point $\text{O}$, due to the arcs $1$ and $3$ are directed normally inwards while field due to arc $2$ is directed normally outwards. Magnetic field due the linear sections of the wires will be zero, because point $\text{O}$ lies on their linear extension.

Magnetic field due to a circular arc of radius $R$, subtending an angle $\theta$ at the center is,

$B=\frac{{\mu }_{0}I}{2R}\left[\frac{\theta }{2\pi }\right]=\frac{{\mu }_{0}I}{4\pi }\times \left(\frac{\theta }{R}\right)$

$\Rightarrow B_1=\frac{{\mu }_{0}I}{4\pi }\left(\frac{\theta }{R}\right)(\otimes )$

$\Rightarrow B_2=\frac{{\mu }_{0}I}{4\pi }\left(\frac{\theta }{2R}\right)(\odot )$

$\Rightarrow B_3=\frac{{\mu }_{0}I}{4\pi }\left(\frac{\theta }{3R}\right)(\otimes )$

The resultant field at point $\text{O}$ is,

$\therefore B_{\text{net}}=\frac{{\mu }_{0}I}{4\pi }[\frac{\theta }{3R}-\frac{\theta }{2R}+\frac{\theta }{R}]$

$\Rightarrow B_{\text{net}}=\frac{5{\mu }_{0}I\theta }{24\pi R}$

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