Question

A conducting wire is in the shape of a regular hexagon which is inscribed inside an imaginary circle of radius $R$. If a current I flows the wire, the magnitude of magnetic field at the centre of the circle is

• A. $\frac{\mu_{0} J}{2 \sqrt{3} \pi R}$
• B. $\frac{\sqrt{3} \mu_{0} J}{2 \pi R}$
• C. $\frac{3 \mu_{0} J}{2 \pi R}$
• D. $\frac{\sqrt{3} \mu_{0} J}{\pi R}$

Answer 1

Answer: (D)

The given situation is shown in the following figure

First we will calculate magnetic field at centre due to current segment $A B$.
$\therefore $ In $ \triangle O A M$
$\sin 60^{\circ}=\frac{O M}{R} $
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{O M}{R}$
$\Rightarrow O M=\frac{\sqrt{3}}{2} R$
$ \Rightarrow r=O M=\frac{\sqrt{3}}{2} R$
$\therefore $ Magnetic field at point $O$ due to current segment $A B$ is given as
$B_{1}=\frac{\mu_{0}}{4 \pi} \cdot \frac{I}{r}\left[\sin 30^{\circ}+\sin 30^{\circ}\right]$
$=\frac{\mu_{0}}{4 \pi} \times \frac{I}{\frac{\sqrt{3}}{2}}\left[\frac{1}{2}+\frac{1}{2}\right]=\frac{\mu_{0} I}{2 \sqrt{3} \pi R}$
[downward ]
$\therefore $ Magnitude to total magnetic field at centre is given as,
$B=6 B_{1}=6 \times \frac{\mu_{0} I}{2 \sqrt{3} \pi R}=\frac{\sqrt{3} \mu_{0} I}{\pi R} $ [downward]