JEE Main 2023 · 30 January · evening · Single correct · +4 / −1
As shown in the figure, a current of 2 A flowing in an equilateral triangle of side 4 √3 cm. The magnetic field at the centroid O of the triangle is
(Neglect the effect of earth's magnetic field)
- (a) 4 √3 × 10⁻⁴ T
- (b) 4 √3 × 10⁻⁵ T
- (c) 3 √3 × 10⁻⁵ T
- (d) √3 × 10⁻⁴ T
Answer and solution
Answer (c) 3 √3 × 10⁻⁵ T
Solution
Physics sir · 4 steps
Treat each side as a finite straight wire. By symmetry all three give equal fields, so find one and triple it.
- Distance to each side. The perpendicular distance from the centroid of an equilateral triangle to any side is r = a/2√3. With a = 4√3 cm, this gives r = 2 cm = 0.02 m.
- Angles at O. Each side subtends 120° at O, split equally by the perpendicular. So each end makes 60° with the perpendicular, and the finite-wire formula uses sin 60° + sin 60° = √3.
B_(side) = (μ₀ I)/(4π r)(sin 60° + sin 60°) = (10⁻⁷ × 2)/0.02 × √3 = √3 × 10⁻⁵ T
- Add the three sides. The current goes round the loop in one sense (anticlockwise in the figure), so all three sides send their field out of the page at O. The fields add: B = 3 × √3 × 10⁻⁵ T.
- So. The field at O is 3√3 × 10⁻⁵ T, which is option (c).
The idea to keep. For a regular polygon loop, find the field of one side with (μ₀ I)/(4π r)(sin α + sin β), where r is the perpendicular distance from the centre. Then multiply by the number of sides.
Answer from the NTA final answer key. Solution by PrepWiser.