Question

The magnetic field at centre of a hexagonal coil of side $l$ carrying $i$ is

• A. $\frac{\sqrt{3}{\mu }_{0}i}{\pi l}$
• B. $\frac{{\mu }_{0}I}{4\pi l}$
• C. $\frac{\pi {\mu }_{0}I}{\sqrt{3}l}$
• D. Zero

Answer 1

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

In the triangle AMO, the angle can be written as,

$\tan \theta =\frac{\frac{l}{2}}{r}$

$\tan {30}^{\circ }=\frac{\frac{l}{2}}{r}$

$r=\frac{\sqrt{3}l}{2}$

The magnetic field due to the side AB is given as,

$B_{AB}=\frac{{\mu }_0}{4\pi }\times \frac{i}{r}\left(\sin \theta +\sin \theta \right)$

$=\frac{{\mu }_0}{4\pi }\times \frac{2i\sin \theta }{r}$

$=\frac{{\mu }_0}{4\pi }\times \frac{2i\sin {30}^{\circ }}{\frac{\sqrt{3}l}{2}}$

$=\frac{{\mu }_0}{4\pi }\times \frac{2i}{\sqrt{3}l}$

Total magnetic field at the centre is given as,

$B_o=6B_{AB}$

$=6\times \frac{{\mu }_0}{4\pi }\times \frac{2i}{\sqrt{3}l}$

$=\frac{{\mu }_{0}i\sqrt{3}}{\pi l}$

Thus, the magnetic field at the centre is $\frac{{\mu }_{0}i\sqrt{3}}{\pi l}$.