JEE Physics 2023 Question with Solution

Given: Current in each side is $2 \, \text{A}$ and the triangle is equilateral with side $4\sqrt{3} \, \text{cm}$.

Find: Magnetic field at the centroid $O$.

From the geometry used in the solution,

$$d \tan 60^\circ = 2\sqrt{3}$$

So,

$$d = 2 \, \text{cm}$$

The magnetic field at the centroid due to the three sides is

$$B = 3 \times \frac{\mu_0 i}{2\pi d} \sin 60^\circ$$

Substituting the given values,

$$B = 3 \times \frac{2 \times 10^{-7} \times 2}{2 \times 10^{-2}} \times \frac{\sqrt{3}}{2}$$

Therefore,

$$B = 3\sqrt{3} \times 10^{-5} \, \text{T}$$

So, the correct option is D.

Given: An equilateral triangle of side $4\sqrt{3} \, \text{cm}$ carries current $2 \, \text{A}$.

Find: The net magnetic field at centroid $O$.

For an equilateral triangle, the perpendicular distance from the centroid to each side is the same. Using the relation shown in the extracted solution,

$$d \tan 60^\circ = 2\sqrt{3}$$

Since $\tan 60^\circ = \sqrt{3}$, we get

$$d \sqrt{3} = 2\sqrt{3}$$ $$d = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m}$$

Now, field due to one side at the centroid is taken as

$$B_1 = \frac{\mu_0 i}{2\pi d} \sin 60^\circ$$

All three sides contribute equally and in the same direction at the centroid, so

$$B = 3B_1 = 3 \times \frac{\mu_0 i}{2\pi d} \sin 60^\circ$$

Substituting $\mu_0/4\pi = 10^{-7}$, $i = 2 \, \text{A}$, $d = 2 \times 10^{-2} \, \text{m}$, and $\sin 60^\circ = \frac{\sqrt{3}}{2}$,

$$B = 3 \times \frac{2 \times 10^{-7} \times 2}{2 \times 10^{-2}} \times \frac{\sqrt{3}}{2}$$ $$B = 3\sqrt{3} \times 10^{-5} \, \text{T}$$

Therefore, the magnetic field at the centroid is $3\sqrt{3} \times 10^{-5} \, \text{T}$, and the correct option is D.