Question

A conducting loop shaped as regular hexagon of side $x$ carrying current $i$ is shown in figure.

If $\text{P}$ is the centre of hexagon, then the magnetic field at point $\text{P}$ is

• A. $\frac{{\mu }_{0}i}{2x}$
• B. zero
• C. $\frac{3{\mu }_{0}i}{32\pi x}$
• D. $\frac{\sqrt{3}{\mu }_{0}i}{\pi x}$

Answer 1

The correct option is D $\frac{\sqrt{3}{\mu }_{0}i}{\pi x}$

The given hexagonal arrangement can be visualized as combination of six current carrying straight wires.

Using relation, for each wire:

$B=\frac{{\mu }_{0}i}{4\pi d}[sin{\varphi }_1+sin{\varphi }_2]$


We can see that $\Delta PAB$ is an equilateral $\Delta$ with each angle ${60}^{\circ }$.

Thus ${\varphi }_1={\varphi }_2=\frac{{60}^{\circ }}{2}={30}^{\circ }$

All the six wire segment are arranged symmetricaly w.r.t the centre $\text{P}$ in the given regular hexagon.

$\Rightarrow B_P=6\times B$

or, $B_P=6\times \frac{{\mu }_{0}i}{4\pi d}[sin{30}^{\circ }+sin{30}^{\circ }]....\left(i\right)$

In $\Delta$ $\text{PAB}$ perpendicular distance of wire $\text{AB}$ from $\text{P}$ is:

$\text{PG}=xsin{60}^{\circ }=\frac{\sqrt{3}x}{2}$

or, $d=\frac{\sqrt{3}x}{2}....\left(ii\right)$

From Eq, (i) and (ii) we have,

$B_P=\frac{6{\mu }_{0}i}{4\pi \left(\frac{\sqrt{3}x}{2}\right)}\times \left[2sin{30}^{\circ }\right]$

$\Rightarrow B_P=\frac{12{\mu }_{0}i}{4\sqrt{3}\pi x}=\frac{3{\mu }_{0}i}{\sqrt{3}\pi x}$

$\therefore B_P=\frac{\sqrt{3}{\mu }_{0}i}{\pi x}$

$\begin{array}{c}\text{Why this question?}\\ \text{From right hand thumb rule or using}\overrightarrow{dl}\times \overrightarrow{r}\text{we infer that the direction of magnetic field at center of hexagon due to each straight wire section is perpendicularly outwards to the plane of hexagon. Thus the net magnetic field is due to sum of all supporting magnetic fields at P.}\end{array}$