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 $AB$.
$\therefore$ In $△OAM$
$\sin 60^{\circ }=\frac{OM}{R}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{OM}{R}$
$\Rightarrow OM=\frac{\sqrt{3}}{2}R$
$\Rightarrow r=OM=\frac{\sqrt{3}}{2}R$
$\therefore$ Magnetic field at point $O$ due to current segment $AB$ 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,
$B=6B_1=6\times \frac{{\mu }_{0}I}{2\sqrt{3}\pi R}=\frac{\sqrt{3}{\mu }_{0}I}{\pi R}$ [downward]