Question

A closed loop is in the form of a regular hexagon of side $a$. If the circuit carries a current $I$, what is the magnetic field at the center of the hexagon ?

• A. $\frac{{\mu }_{0}I}{\sqrt{3}\pi a}$
• B. $\frac{{\mu }_{0}I}{\pi a}$
• C. $\frac{\sqrt{3}{\mu }_{0}I}{\pi a}$
• D. $\frac{\sqrt{3}{\mu }_{0}I}{2\pi a}$

Answer 1

The correct option is C $\frac{\sqrt{3}{\mu }_{0}I}{\pi a}$


Since, the above diagram is a regular hexagon, so $\Delta \text{ABO}$ will be an equilateral triangle.

So, the height of an equilateral triangle is given by

$CO=\frac{\sqrt{3}}{2}a$

Magnetic field due to wire $AB$ at centre $O$ is given by

$B=\frac{{\mu }_0}{4\pi }\frac{I}{\left(CO\right)}[\sin {\theta }_1+\sin {\theta }_2]$

$B=\frac{{\mu }_0}{4\pi }\frac{I}{\left(\sqrt{3}a/2\right)}[\sin {30}^{\circ }+\sin {30}^{\circ }]$

$B=\frac{{\mu }_0}{2\sqrt{3}\pi }\frac{I}{a}$

From right- hand thumb rule, we can conclude that the direction of magnetic field at the centre $O$ due to all six sides of hexagon will be same and that will be perpendicular outward of the plane of hexagon.

Thus, the net magnetic field for all elements is given by,

$B_{net}=6B=6\times \frac{{\mu }_0}{2\sqrt{3}\pi }\frac{I}{a}$

$B_{net}=\frac{\sqrt{3}{\mu }_{0}I}{\pi a}$

So, option $\left(c\right)$ is the right answer.