A flat circular loop carrying a current of 2.0 A is in a $\vec { B }$ field of 3.5 T.The loop has an area of 0.12 m² and its plane is oriented at a 37° angle to the $\vec { B }$ field.What is the magnitude of the magnetic torque exerted on the loop?

Question

Quizplus · Accepted Answer

The formula for magnetic torque is:

$\vec \tau = \vec {N} \times \vec{B}$

where:
- $\vec {N}$ is the normal vector to the area of the loop (in the same direction as the current)
- $\vec{B}$ is the magnetic field vector

The magnitude of the torque is:

|$\vec \tau$| = |$\vec {N}$| |$\vec{B}$|sinθ

where θ is the angle between $\vec {N}$ and $\vec{B}$.

In this case, the area of the loop is given by A = 0.12 m^2, so |$\vec {N}$| = A = 0.12 m^2.

The angle between $\vec {N}$ and $\vec{B}$ is 37°, so sinθ = sin37° = 0.6.

Therefore, the magnitude of the torque is:

|$\vec \tau$| = |$\vec {N}$| |$\vec{B}$|sinθ = 0.12 m^2 × 2.0 A × 3.5 T × 0.6 = 0.672 N ∙ m ≈ 0.67 N ∙ m

So the best choice is C.

A Flat Circular Loop Carrying a Current of 2 B⃗\vec { B }B

A Flat Circular Loop Carrying a Current of 2 $\vec { B }$