Question

The current passing through a conducting loop in the form of equilateral triangle of side $4\sqrt{3}$ cm is 2 A. The magnetic field at its centroid is $\alpha \times 10^{-5}$ T. The value of $\alpha$ is ___. (Given : $\mu_0 = 4\pi \times 10^{-7}$ SI units)

• A. $3\sqrt{3}$
• B. $2\sqrt{3}$
• C. $\sqrt{3}$
• D. $\frac{\sqrt{3}}{2}$

Hint:

The magnetic field at the center of a regular polygon with $n$ sides is $n \times$ field of one side. $B = \frac{n \mu_0 I}{2\pi R} \tan(\pi/n)$ (where R is circumradius).

Answer 1

The Correct Option is A

The distance from the centroid to the side of an equilateral triangle is $d = \frac{a}{2\sqrt{3}}$.
Given $a = 4\sqrt{3} \text{ cm}$. $d = \frac{4\sqrt{3}}{2\sqrt{3}} = 2 \text{ cm} = 0.02 \text{ m}$.
The magnetic field due to one side at the centroid is given by $B_1 = \frac{\mu_0 I}{4\pi d} (\sin \theta_1 + \sin \theta_2)$.
For an equilateral triangle, the ends of a side subtend angles $60^\circ$ and $60^\circ$ at the centroid.
$B_1 = \frac{4\pi \times 10^{-7} \times 2}{4\pi \times 0.02} (2 \sin 60^\circ) = \frac{2 \times 10^{-7}}{0.02} (\sqrt{3}) = \sqrt{3} \times 10^{-5} \text{ T}$.
The total field is 3 times the field of one side: $B_{net} = 3 \times B_1 = 3\sqrt{3} \times 10^{-5} \text{ T}$.
Comparing with $\alpha \times 10^{-5}$, we get $\alpha = 3\sqrt{3}$.