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: 
$\boldsymbol{	au} = \boldsymbol{\mu} 	imes \boldsymbol{B}$ 
where $\boldsymbol{\mu}$ is the magnetic dipole moment of the loop given by $I\cdot \boldsymbol{A}$ and $\boldsymbol{A}$ is the area vector of the loop. 
For a circular loop, the magnetic dipole moment can be written as $\boldsymbol{\mu} = I\pi r^2 \boldsymbol{n}$ where $\boldsymbol{n}$ is the unit normal vector to the loop. 
In this problem, the magnetic dipole moment is: 
$\boldsymbol{\mu} = (3.0	ext{ A})(\pi (0.50	ext{ m})^2) (11\boldsymbol{e}_x/3) = 2.915\boldsymbol{e}_x$ A$\cdot$m$^2$. 
The torque on the loop is then given by: 
$\boldsymbol{	au} = \boldsymbol{\mu} 	imes \boldsymbol{B} = (2.915\boldsymbol{e}_x 	ext{ A}\cdot 	ext{m}^2) 	imes (1.41\boldsymbol{e}_z 	ext{ T}) = 4.11 \boldsymbol{e}_y$ N$\cdot$m. 
Therefore, the x component of the torque is 0 and the y component is 4.11 N$\cdot$m. The closest answer choice to this value is (D) 9.4 N$\cdot$m.

A Circular Loop (Radius = 0