JEE Main Physics Class 12 Moving Charges and Magnetism Part 2 Questions

Side of equilateral triangular coil, a = 10 cm = 0.1 m
Horizontal magnetic field, B=20‌mT=20×10−3T
Current in coil, i = 0.2 A
Torque acting on coil, τ=√x×10−5N−m
Torque acting on coil can be calculated as
τ = MBsin θ Here, M is pole strength and B is magnetic field.Magnetic pole strength, M = iA
τ=i(

√3

4

a2)B‌sin‌θ [∵A=

√3

4

a2. (for equilateral triangle)]
=0.2×

√3

4

×(0.1)2×20×10−3×sin‌90° [∵ A ⊥ B]
=√3×10−5N−m
After comparing obtained value with the given value, x = 3.