The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is : [Take \mu 0 = 4 \pi × 10–7 NA–2]

Magnetics PYQ — JEE Main 2019 · 10 Apr · Shift 2 · Q47 with Solution

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Correct answer: B

For a finite straight wire, the magnetic field at a point at perpendicular distance $r$ is

B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)

At the centre of an equilateral triangle, the perpendicular distance from the centre to each side is the inradius:

r = \frac{a\sqrt{3}}{6}

Given $a=1\,\text{m}$ ,

r = \frac{\sqrt{3}}{6}\,\text{m}

For one side, the centre lies on the perpendicular bisector of the side, so the two angles are equal:

\theta_1 = \theta_2 = \theta

Now,

\tan\theta = \frac{(a/2)}{r} = \frac{1/2}{\sqrt{3}/6} = \sqrt{3}

Hence,

\theta = 60^\circ

So magnetic field due to one side is

B_1 = \frac{\mu_0 I}{4\pi r}(\sin 60^\circ + \sin 60^\circ) = \frac{\mu_0 I}{4\pi r}(2\cdot \tfrac{\sqrt{3}}{2}) = \frac{\mu_0 I\sqrt{3}}{4\pi r}

Substitute $r=\frac{\sqrt{3}}{6}$ :

B_1 = \frac{\mu_0 I\sqrt{3}}{4\pi \cdot (\sqrt{3}/6)} = \frac{6\mu_0 I}{4\pi} = \frac{3\mu_0 I}{2\pi}

All three sides produce magnetic field at the centre in the same direction, so

B = 3B_1 = 3\cdot \frac{3\mu_0 I}{2\pi} = \frac{9\mu_0 I}{2\pi}

Now substitute $\mu_0 = 4\pi \times 10^{-7}$ and $I=10\,\text{A}$ :

B = \frac{9}{2\pi}(4\pi \times 10^{-7})(10)

B = 18 \times 10^{-6}\,\text{T}

B = 18\,\mu\text{T}

Therefore, the correct answer is Option B.