A wire shaped in a regular hexagon of side 2 cm carries a current of 4A. The magnetic field at the centre of hexagon is

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Aicharya · Accepted Answer

$\mathrm{r}=\frac{\frac{\mathrm{a}}{2}}{\tan 30^{\circ}}=\frac{\sqrt{3} \mathrm{a}}{2}=\sqrt{3} \mathrm{~cm}$ $\therefore$ Magnetic field at centre 0 , $\begin{aligned} & B=6 \times \frac{\mu_0}{4 \pi} \cdot \frac{I}{r} \cdot\left(2 \sin 30^{\circ}\right) \\ & =6 \times 10^{-7} \times \frac{4}{\sqrt{3} \times 10^{-2}} \times 2 \times \frac{1}{2} \\ & =8 \sqrt{3} \times 10^{-5} \mathrm{~T} \end{aligned}$

A wire shaped in a regular hexagon of side 2 cm carries a current of 4A. The magnetic field at the centre of…

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