Q59 / 75 · JEE Main 2026 (23 January Shift 2) · Physics
The current passing through a conducting loop in the form of equilateral triangle of side $4 \sqrt{3} \mathrm{~cm}$ is 2 A. The magnetic field at its centroid is $\alpha \times 10^{-5} \mathrm{~T}$. The value of $\alpha$ is $\_\_\_\_$.
(Given : $\mu_{\mathrm{o}}=4 \pi \times 10^{-7}$ SI units)
(Given : $\mu_{\mathrm{o}}=4 \pi \times 10^{-7}$ SI units)
- A$2 \sqrt{3}$
- B$\frac{\sqrt{3}}{2}$
- Correct answerC$3 \sqrt{3}$
- D$\sqrt{3}$
Solution
For an equilateral triangle of side $a$, the distance of the centroid from any side is $r = \frac{a}{2\sqrt{3}}$.
Given side $a = 4\sqrt{3}$ cm $= 4\sqrt{3} \times 10^{-2}$ m.
So, $r = \frac{4\sqrt{3} \times 10^{-2}}{2\sqrt{3}} = 2 \times 10^{-2}$ m.
The magnetic field at the centroid due to one side of the triangle is $B_1 = \frac{\mu_0 I}{4\pi r} (\sin \theta_1 + \sin \theta_2)$.
For an equilateral triangle, $\theta_1 = \theta_2 = 60^\circ$.
$B_1 = \frac{10^{-7} \times 2}{2 \times 10^{-2}} (\sin 60^\circ + \sin 60^\circ) = 10^{-5} \times (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) = \sqrt{3} \times 10^{-5}$ T.
The total magnetic field at the centroid due to all three sides is $B = 3B_1 = 3\sqrt{3} \times 10^{-5}$ T.
Comparing with $B = \alpha \times 10^{-5}$ T, we get $\alpha = 3\sqrt{3}$.
Given side $a = 4\sqrt{3}$ cm $= 4\sqrt{3} \times 10^{-2}$ m.
So, $r = \frac{4\sqrt{3} \times 10^{-2}}{2\sqrt{3}} = 2 \times 10^{-2}$ m.
The magnetic field at the centroid due to one side of the triangle is $B_1 = \frac{\mu_0 I}{4\pi r} (\sin \theta_1 + \sin \theta_2)$.
For an equilateral triangle, $\theta_1 = \theta_2 = 60^\circ$.
$B_1 = \frac{10^{-7} \times 2}{2 \times 10^{-2}} (\sin 60^\circ + \sin 60^\circ) = 10^{-5} \times (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) = \sqrt{3} \times 10^{-5}$ T.
The total magnetic field at the centroid due to all three sides is $B = 3B_1 = 3\sqrt{3} \times 10^{-5}$ T.
Comparing with $B = \alpha \times 10^{-5}$ T, we get $\alpha = 3\sqrt{3}$.
Hence, the correct answer is option: 'C'.