Physics - Magnetism Question with Solution | TestHub

The magnetic field at the center is the sum of fields from $n$ sides. Each side acts as a finite wire segment.

The magnetic field due to a finite straight current-carrying wire segment at a point is given by:

$$B=\frac{{\mu }_{0}I}{4\pi r}(\sin {\theta }_1+\sin {\theta }_2)$$

where $r$ is the perpendicular distance from the wire to the point, and ${\theta }_1$, ${\theta }_2$ are the angles subtended by the ends of the wire segment with the perpendicular.

For a regular $n$-sided polygon inscribed in a circle of radius $a$, the center of the circle is equidistant from all sides.

Let's consider one side of the polygon.

The angle subtended by each side at the center is $\frac{2\pi }{n}$.

The angle subtended by half a side at the center is $\frac{\pi }{n}$.

The perpendicular distance from the center to a side, $r$, can be found using trigonometry:

$r=a\cos \left(\frac{\pi }{n}\right)$

For each side, the angles ${\theta }_1$ and ${\theta }_2$ are equal to $\frac{\pi }{n}$.

So, $\sin {\theta }_1=\sin {\theta }_2=\sin \left(\frac{\pi }{n}\right)$.

The magnetic field due to one side at the center is:

$$B_{side}=\frac{{\mu }_{0}i}{4\pi r}\left(\sin \left(\frac{\pi }{n}\right)+\sin \left(\frac{\pi }{n}\right)\right)$$

$$B_{side}=\frac{{\mu }_{0}i}{4\pi a\cos \left(\frac{\pi }{n}\right)}\left(2\sin \left(\frac{\pi }{n}\right)\right)$$

$$B_{side}=\frac{{\mu }_{0}i}{2\pi a}\frac{\sin \left(\frac{\pi }{n}\right)}{\cos \left(\frac{\pi }{n}\right)}$$

$$B_{side}=\frac{{\mu }_{0}i}{2\pi a}\tan \left(\frac{\pi }{n}\right)$$

Since there are $n$ such sides, and the magnetic fields due to all sides point in the same direction (perpendicular to the plane of the polygon, at the center), the total magnetic flux density at the center is:

$$B_{total}=n\times B_{side}$$

$$B_{total}=n\times \frac{{\mu }_{0}i}{2\pi a}\tan \left(\frac{\pi }{n}\right)$$

$$B_{total}=\frac{{\mu }_{0}ni}{2\pi a}\tan \left(\frac{\pi }{n}\right)$$

The final answer is $\boxed{\frac{{\mu }_{0}n\mathrm{i}}{2\pi \mathrm{a}}\tan \frac{\pi }{\mathrm{n}}}$.