A circular loop (radius = 0.50 m) carries a current of 3.0 A and has unit normal vector of ()/3.What is the x component of the torque on this loop when it is placed in a uniform magnetic field of ()T?
A)4.7 N . m
B)3.1 N .m
C)19 N . m
D)9.4 N.m
E)12 N . m

Question

A circular loop (radius = 0.50 m) carries a current of 3.0 A and has unit normal vector of ()/3.What is the x component of the torque on this loop when it is placed in a uniform magnetic field of ()T?  
A)4.7 N . m 
B)3.1 N .m 
C)19 N . m 
D)9.4 N.m 
E)12 N . m

Quizplus · Accepted Answer

The torque on a current loop in a magnetic field is given by the formula τ = IABsinθ, where I is the current, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the magnetic field and the normal to the loop. Since the loop is circular, its area is given by A = πr^2 = π(0.5 m)^2 = 0.785 m^2. The angle between the magnetic field and the normal to the loop is 60 degrees, since the unit normal vector is ()/3 and the magnetic field is uniform. Therefore, sinθ = sin(60°) = √3/2. Substituting the given values, we get τ = (3.0 A)(0.785 m^2)(0.5 T)(√3/2) = 9.4 N m. The x-component of the torque is given by τx = τsinθ = (9.4 N m)(√3/2) ≈ 8.14 N m, which is closest to choice D.

A Circular Loop (Radius = 0