Chapter 20, Problem 64P

(a)

To determine

The magnetic force on side 2 and 4 at the instant shown in Figure 20.7.

Answer to Problem 64P

The magnetic force on side 2 at the instant shown in Figure 20.7 is $\underset{\_}{{\overrightarrow{F}}_2=\frac{\omega B^{2}AL}{R}\sin \omega t,\text{down}}$ and that on side 4 is $\underset{\_}{{\overrightarrow{F}}_4=\frac{\omega B^{2}AL}{R}\sin \omega t,\text{up}}$.

Explanation of Solution

In side 2, current flows into the page and in side 4, current flows out of the page.

Given that the current in the system is,

  $I\left(t\right)=\frac{\omega BA}{R}\sin \omega t$                                                                                                      (I)

Write the expression for the magnetic force due to the current carrying loop.

  $\overrightarrow{F}=I\overrightarrow{L}\times \overrightarrow{B}$                                                                                                                  (II)

Here, $\overrightarrow{F}$ is the magnetic force, $I$ is the current, $\overrightarrow{L}$ is the side length of the loop, and $\overrightarrow{B}$ is the magnetic field.

$\overrightarrow{L}$ and $\overrightarrow{B}$ are always perpendicular for both sides. Thus, the magnitude of the magnetic force on each side is same and is obtained as,

  $F=ILB$ (III)

Use equation (I) in (III).

  $\begin{array}{c}F=\frac{\omega BA}{R}LB\sin \omega t\\ =\frac{\omega B^{2}AL}{R}\sin \omega t\end{array}$

The direction of the magnetic force is determined using the right-hand rule, and applying right hand rule yields that the magnetic force on side 2 is directed downward and that on side 4 is directed upward.

Conclusion:

Therefore, the magnetic force on side 2 at the instant shown in Figure 20.7 is $\underset{\_}{{\overrightarrow{F}}_2=\frac{\omega B^{2}AL}{R}\sin \omega t,\text{down}}$ and that on side 4 is $\underset{\_}{{\overrightarrow{F}}_4=\frac{\omega B^{2}AL}{R}\sin \omega t,\text{up}}$.

(b)

To determine

The reason for which the magnetic forces on side 1 and 3 do not cause a torque about the axis of rotation.

Answer to Problem 64P

The magnetic forces on side 1 and 3 are always parallel to the axis of rotation, and hence they cannot produce a torque about the axis.

Explanation of Solution

If the forces acting on a body is parallel to the axis of rotation of the body, the forces cannot produce a torque on the body. In the given system, the magnetic forces on side 1 and 3 are parallel to the axis of rotation. Hence, they cannot cause a torque about the axis.

Conclusion:

Therefore, the magnetic forces on side 1 and 3 are always parallel to the axis of rotation, and hence they cannot produce a torque about the axis.

(c)

To determine

The torque on the loop about its axis of rotation.

Answer to Problem 64P

The torque on the loop about its axis of rotation is $\underset{\_}{\overrightarrow{\tau }=\frac{2\omega r{B}^{2}AL}{R}{\sin }^2\omega t,\text{counterclockwise}}$.

Explanation of Solution

The diagram depicting the forces on the loop is shown in Figure 1.

Write the expression for the total torque on the loop using the diagram.

  ${\displaystyle \sum \tau }=\left(F_{\text{2}}\sin \theta \right)r+\left(F_{\text{4}}\sin \theta \right)r$                                                                              (IV)

Here, $\tau$ is the torque, $\theta$ is the angular displacement, and $r$ is the half of side length of the loop.

The magnitude of magnetic force $F$ is same on side 2 and 4 and thus replace $F_2$ and $F_4$ by $F$ in equation (IV).

  $\begin{array}{c}{\displaystyle \sum \tau }=\left(F\sin \theta \right)r+\left(F\sin \theta \right)r\\ =2Fr\sin \theta \end{array}$                                                                                (V)

Write the expression for the angular displacement.

  $\theta =\omega t$                                                                                                                   (VI)

Here, $\omega$ is the angular speed, and $t$ is the time.

From part (a) the magnetic force is obtained as,

  $F=\frac{\omega B^{2}AL}{R}\sin \omega t$                                                                                               (VII)

Use equation (VI) and (VII) in(V).

  $\begin{array}{c}{\displaystyle \sum \tau }=2\left(\frac{\omega B^{2}AL}{R}\sin \omega t\right)r\sin \omega t\\ =\frac{2\omega r{B}^{2}AL}{R}{\sin }^2\omega t\end{array}$

The direction of forces indicates that the torque on the loop is in the counter clockwise direction.

Conclusion:

Therefore, the torque on the loop about its axis of rotation is $\underset{\_}{\overrightarrow{\tau }=\frac{2\omega r{B}^{2}AL}{R}{\sin }^2\omega t,\text{counterclockwise}}$.

(d)

To determine

Whether the magnetic torque make the loop increase or decrease its angular velocity, in the absence of other torques.

Answer to Problem 64P

In the absence of other torques the magnetic torque make the loop decrease its angular velocity.

Explanation of Solution

In the given system, the loop has the angular velocity in the clockwise direction whereas the torque exerted by the magnetic force is obtained as in the counterclockwise direction. Since the direction of magnetic torque is opposite to the direction of the angular velocity, the magnetic torque would tend to decrease the angular speed of the loop.

Conclusion:

Therefore, in the absence of other torques the magnetic torque make the loop decrease its angular velocity.