A flat rectangular loop of wire is placed between the poles of a magnet,as shown in the figure.It has dimensions w = 0.60 m and L = 1.0 m,and carries a current I = 2.0 A in the direction shown.The $\vec { B }$ field due to the magnet is uniform and of magnitude 0.80 T.The loop rotates in the magnetic field and at one point the plane of the loop makes a 30° angle with the $\vec { B }$ field vector.At that instant,what is the magnitude of the torque exerted on the wire by the magnetic field?

Question

Quizplus · Accepted Answer

The torque, given by $\vec{	au} = \vec{r} 	imes \vec{F}$, where $\vec{r}$ is the position vector and $\vec{F}$ is the force acting on the loop.
At an angle of $30^\circ$, the component of the magnetic field perpendicular to the loop is $B\sin{30^\circ} = 0.4$ T. The area of the loop is $A = wL = 0.6 	imes 1.0 = 0.6$ m$^2$. The magnitude of the force on each side of the loop is $F = BIL = 0.4 	imes 2.0 	imes 0.6 = 0.48$ N.

The torque is given by:
$$
\begin{aligned}
	au &= r F \sin 	heta \
&= \left(\frac{L}{2}ight) (2F) \sin(30^\circ) \
&= 0.5 	imes 2 	imes 0.48 	imes 0.5 \
&= 0.48 	ext{ Nm}
\end{aligned}
$$

Therefore, the best choice is C.

A Flat Rectangular Loop of Wire Is Placed Between the Poles