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 magnetic torque on a current loop can be given by the formula τ = NIABsinθ, where N is the number of turns, I is the current, A is the area of the loop and θ is the angle between the normal to the loop and the magnetic field.

Here, N = 1 (one turn), I = 3.0 A, A = πr^2 = 0.785 m^2, and θ = 0 degrees (since the magnetic field is parallel to the normal to the loop).

The x-component of the magnetic field is given as 0.60 T. Therefore, the magnetic torque is given by:

τ = (1)(3.0 A)(0.785 m^2)(0.60 T)sin(0°) = 9.4 N⋅m

Hence, the correct answer is D.

A Circular Loop (Radius = 0