A current carrying circular loop of radius 2 cm with unit normal n = k + i 2 is placed in a magnetic field, B = B₀(3 i +2 k ) . If B₀ = 4 10⁻³ T and current I=100 2 A, the torque experienced by the loop is ________ Wb·A. ( =3.14 )

Step-by-step solution

The magnetic dipole moment of the current carrying loop is given by M = I A n . Given radius r = 2 cm = 2 10⁻² m, the area of the loop is: A = r^2 = (2 10⁻²)^2 = 4 10⁻⁴ m ^2 Substituting the given values I = 100 2 A and n = i + k 2 : M = (100 2 ) (4 10⁻⁴) ( i + k 2 ) M = 4 10⁻² ( i + k ) A m ^2 The torque experienced by the loop in the magnetic field is = M B . Given B = 4 10⁻³ (3 i +2 k ) T, we have: = [4 10⁻² ( i + k )] [4 10⁻³ (3 i +2 k )] = 16 10⁻⁵ [( i + k ) (3 i +2 k )] Evaluating the cross product: ( i + k )