Question

The magnetic field induction at the centre of a regular polygon having n sides is (The polygon is inscribed in a circle of radius 'a' metre. The side of the polygon carries a current of I amp)(Current is clockwise)

• A. $\frac{2In}{a} \tan \pi/n \times 10^{-7}$ tesla(outward)
• B. $\frac{2In}{a} \tan \pi/n \times 10^{-7}$ tesla(inward)
• C. $\frac{In}{a} \tan \pi/n \times 10^{-7}$ tesla(inward)
• D. $\frac{In}{a} \tan \pi/n \times 10^{-7}$ tesla (outward)

Answer 1

Answer: (B)

$\frac{2In}{a} \tan \pi/n \times 10^{-7}$ tesla (inward)

Answer 2

The magnetic field at the center of a regular polygon with 'n' sides, inscribed in a circle of radius 'a', carrying current 'I', is the sum of magnetic fields due to each side.

1. Perpendicular distance from center to a side: $d = a \cos(\pi/n)$.

2. Angles subtended by half a side at the center: $\theta_1 = \theta_2 = \pi/n$.

3. Magnetic field due to one side: $B_{side} = \frac{\mu_0 I}{4\pi d} (\sin\theta_1 + \sin\theta_2) = \frac{\mu_0 I}{4\pi a \cos(\pi/n)} (2\sin(\pi/n)) = \frac{\mu_0 I}{2\pi a} \tan(\pi/n)$.

4. Total magnetic field: $B_{total} = n \times B_{side} = n \frac{\mu_0 I}{2\pi a} \tan(\pi/n)$.

5. Substitute $\mu_0 = 4\pi \times 10^{-7}$: $B_{total} = \frac{2In}{a} \tan(\pi/n) \times 10^{-7}$ Tesla.

6. Direction: For clockwise current, by Right-Hand Thumb Rule, the field at the center is inward.