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A conducting wire is in the shape of a regular hexagon which is inscribed inside an imaginary circle of radius $R$. If a current $I$ flows the wire, the magnitude of magnetic field at the centre of the circle is
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The correct answer is:
$\frac{\sqrt{3} \mu_0 J}{\pi R}$
The given situation is shown in the following figure
First we will calculate magnetic field at centre due to current segment $A B$.
$\therefore \text { In } \triangle O A M \text {, }$
$\sin 60^{\circ}=\frac{O M}{R} \Rightarrow \frac{\sqrt{3}}{2}=\frac{O M}{R}$
$\Rightarrow \quad O M=\frac{\sqrt{3}}{2} R \Rightarrow \quad r=O M=\frac{\sqrt{3}}{2} R$
$\therefore$ Magnetic field at point $O$ due to current segment $A B$ is given as
$B_1=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left[\sin 30^{\circ}+\sin 30^{\circ}\right]$
$=\frac{\mu_0}{4 \pi} \times \frac{I}{\frac{\sqrt{3}}{2} R}\left[\frac{1}{2}+\frac{1}{2}\right]=\frac{\mu_0 I}{2 \sqrt{3} \pi R} \quad$ [downward]
$\therefore$ Magnitude to total magnetic field at centre is given as,
$B=6 B_1=6 \times \frac{\mu_0 I}{2 \sqrt{3} \pi R}=\frac{\sqrt{3} \mu_0 I}{\pi R} \quad \text { [downward] }$
First we will calculate magnetic field at centre due to current segment $A B$.
$\therefore \text { In } \triangle O A M \text {, }$
$\sin 60^{\circ}=\frac{O M}{R} \Rightarrow \frac{\sqrt{3}}{2}=\frac{O M}{R}$
$\Rightarrow \quad O M=\frac{\sqrt{3}}{2} R \Rightarrow \quad r=O M=\frac{\sqrt{3}}{2} R$
$\therefore$ Magnetic field at point $O$ due to current segment $A B$ is given as
$B_1=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left[\sin 30^{\circ}+\sin 30^{\circ}\right]$
$=\frac{\mu_0}{4 \pi} \times \frac{I}{\frac{\sqrt{3}}{2} R}\left[\frac{1}{2}+\frac{1}{2}\right]=\frac{\mu_0 I}{2 \sqrt{3} \pi R} \quad$ [downward]
$\therefore$ Magnitude to total magnetic field at centre is given as,
$B=6 B_1=6 \times \frac{\mu_0 I}{2 \sqrt{3} \pi R}=\frac{\sqrt{3} \mu_0 I}{\pi R} \quad \text { [downward] }$
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