JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Magnetic Effects of Current

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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

μ_{0} = 4 π \times 1 0^{- 7} NA^{- 2}

]

(A) $9 μ T$
(B) $1 μ T$
(C) $3 μ T$
(D) $18 μ T$ ✓ Correct

Answer: (D)

Solution

The centroid of an equilateral triangle of side

a = 1 m

is a perpendicular distance

d = \frac{a}{2 3}

from each side, and each side subtends

6 0^{\circ}

at the centroid on either side of the foot of the perpendicular. One side gives

B_{1} = \frac{μ _{0} I}{4 π d} (2 sin 6 0^{\circ}) = \frac{6 μ _{0} I}{4 π}

, so the three sides together give

B = \frac{18 μ _{0} I}{4 π} = 18 \times 1 0^{- 7} \times 10 = 18 μ T

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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