Question

A current is flowing in a hexagonal coil of side d. The magnetic induction at the center of the coil will be:

• A. $\frac{3\sqrt{3}\mu i}{\pi d}$
• B. $\frac{\sqrt{3}\mu i}{2\pi d}$
• C. $\frac{\mu i}{3\sqrt{3}\pi d}$
• D. $\frac{\sqrt{3}\mu i}{\pi d}$

Answer 1

The correct option is D $\frac{\sqrt{3}\mu i}{\pi d}$

Let us consider one side of regular hexagon QR as shown in the diagram.
We want to calculate magnetic field at a point O, center of hexagon, such that OQR is an equilateral triangle.
Taking a small element of length $dl$ along QR at a distance of $r$ from O then according to Biot-Savart law, we have
$dB=\frac{\mu idl\sin \theta }{4\pi r^2}$
Also, $\angle R={60}^0$ and $\angle PQO={120}^0$ where P is the point in the line QR above Q.
The resultant field $B$ at point O is obtained by integrating the field due to all the elements from Q to R.
Now,
$r=scosec\theta$ where $s$ is the distance between center O and mid point of QR.
$l=s\cot \theta$ where $l$ is the length of conductor from center of QR to element $dl$
Or, $dl=-s{cosec}^2\theta d\theta$
From Biot Savart law, we have
$dB=\frac{-\mu iscose{c}^{2}d\theta \sin \theta }{4\pi s^{2}cose{c}^2\theta }$
$=\frac{-\mu i\sin \theta d\theta }{4\pi s}$
Integrating above, we get
$B={\int }_{{120}^0}^{{60}^0}\frac{-\mu i\sin \theta d\theta }{4\pi s}$
$B=\frac{\mu i(\cos {60}^0-\cos {120}^0)}{4\pi s}$ ........................ (1)
Also, $s=\frac{\sqrt{3}d}{2}$ ............................... (2)
$\therefore$ from (1) and (2),
$B=\frac{\mu i}{2\sqrt{3}\pi d}$
$\therefore$ for 6 sides, we have
$B^{\prime }=6\frac{\mu i}{2\sqrt{3}\pi d}$$=\sqrt{3}\frac{\mu i}{\pi d}$