JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Magnetic Effects of Current

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A current of 1.5 A is flowing through a triangle, of side 9 cm each. The magnetic field at the centroid of the triangle is :
(Assume that the current is flowing in the clockwise direction.)

(A) $23 \times 1 0^{- 7} T$ , outside the plane of triangle
(B) $23 \times 1 0^{- 5} T$ , inside the plane of triangle
(C) $3 \times 1 0^{- 5} T$ , inside the plane of triangle✓ Correct
(D) $3 \times 1 0^{- 7} T$ , outside the plane of triangle

Answer: (C)

Solution

The field at the centroid is the sum of the fields produced by the three sides, which are equal and add in the same sense. For an equilateral triangle of side

a

the perpendicular distance from the centroid to a side is

r = \frac{a}{2 3}

, and each side subtends a half-angle of

6 0^{\circ}

at the centroid, so one side contributes

B_{1} = \frac{μ _{0} I}{4 π r} (2 sin 6 0^{\circ}) = \frac{μ _{0} I}{4 π} \cdot \frac{2 3 \cdot 3}{a} = \frac{3 μ _{0} I}{2 π a} .

The three sides together give

B = 3 B_{1} = \frac{9 μ _{0} I}{2 π a} = \frac{9 ( 4 π \times 1 0 ^{- 7} ) ( 1.5 )}{2 π ( 0.09 )} = 3 \times 1 0^{- 5} T .

For a clockwise current the field points into the plane of the triangle, so the field at the centroid is

3 \times 1 0^{- 5}

T directed into the plane.

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Magnetic Effects of Current

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