A circular loop (radius = 0.50 m)carries a current of 3.0 A and has unit normal vector of ( $2 \tilde { \mathbf { i } } - \tilde { \mathbf { j } } + 2 \tilde { \mathbf { k } }$ )/3.What is the x component of the torque on this loop when it is placed in a uniform magnetic field of ( $2 \tilde { \mathbf { i } } - 6 \tilde { \mathbf { j } }$ )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

Quizplus · Accepted Answer

The torque on a current loop in a magnetic field is given by the equation

τ = NIABsinθ

where N is the number of turns in the loop, I is the current, A is the area of the loop, B is the magnetic field, and θ is the angle between the normal vector of the loop and the magnetic field.

In this problem, N = 1, I = 3.0 A, A = πr2 = 0.79 m2, B = (1.1, -0.8, 0) T (given in vector form), and θ = 90° because the magnetic field is perpendicular to the normal vector of the loop.

Taking the cross product of the normal vector of the loop and the magnetic field, we get 
( 11ea99f6_d6fa_16be_8bb6_91605b2ed2bb_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11 ) × ( 11ea99f6_d6fa_16bf_8bb6_976c16b58016_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11 ) 
= (0, 0, 1.9) Tm

The x component of this torque is τx = NIABsinθ = (1)(3.0 A)(0.79 m2)(1.9) sin 90° = 9.4 Nm.

Therefore, the answer is choice D, 9.4 N F1F1F1S1F1F1F10 m.

A Circular Loop (Radius = 0 2i~−j~+2k~2 \tilde { \mathbf { i } } - \tilde { \mathbf { j } } + 2 \tilde { \mathbf { k } }2i~−j~​+2k~

A Circular Loop (Radius = 0 $2 \tilde { \mathbf { i } } - \tilde { \mathbf { j } } + 2 \tilde { \mathbf { k } }$