Chapter 22, Problem 28P

(a)

To determine

The magnitude of the torque.

Answer to Problem 28P

The magnitude of the torque is $\underset{\_}{\text{6}\text{.40}\times {\text{10}}^{-4}\text{N}\cdot \text{m}}$

Explanation of Solution

Write the formula for torque on a rotor.    

  $\tau =INAB\sin \theta$        (I)

Here, $\tau$ is the torque, $I$ is current on the particle, $N$ is number of turns, $\theta$ is the angle between $I$ and $B$, and $B$ is the magnetic field.

Conclusion:

Substitute $80$  for $N$ , $\text{10}\text{.0}\times {\text{10}}^{-3}\text{A}$  for $I$  , $\text{0}\text{.025}\text{m}$  for $A$  , $\text{0}\text{.0400m}$  for $A$  , $\text{0}\text{.800T}$  and for $B$, $90°$ for $\theta$  in equation (I),

  $\begin{array}{c}\tau =\left(80\right)\left(\text{10}\text{.0}\times {\text{10}}^{-3}\text{A}\right)\left(\text{0}\text{.025}\text{m}\right)\left(\text{0}\text{.0400m}\right)\sin 90°\\ =\text{6}\text{.40}\times {\text{10}}^{-4}\text{N}\cdot \text{m}\end{array}$

The magnitude of the torque is $\underset{\_}{\text{6}\text{.40}\times {\text{10}}^{-4}\text{N}\cdot \text{m}}$

(b)

To determine

The peak power output.

Answer to Problem 28P

The peak power output is $\underset{\_}{\text{0}\text{.241 W}}$

Explanation of Solution

Write the formula for power on a motor.    

  $P_{\max }={\tau }_{\max }\omega$        (I)

Here, $P_{\max }$ is the power, ${\tau }_{\max }$ is torque, and $\omega$ is the angular velocity.

Conclusion:

Substitute $\text{6}\text{.40}\times {\text{10}}^{-3}\text{N}\cdot \text{m}$  for ${\tau }_{\max }$  , $\text{3 600 rev/min}$  for $\omega$  equation (I),

  $\begin{array}{c}{P}_{\max }=\left(\text{6}\text{.40}\times {\text{10}}^{-3}\text{N}\cdot \text{m}\right)\left(\text{3600 rev/min}\right)\left(\frac{2\pi rad}{1rev}\right)\left(\frac{1\min }{60s}\right)\\ =\text{0}\text{.241 W}\end{array}$

The peak power output is $\underset{\_}{\text{0}\text{.241 W}}$

(c)

To determine

The amount of work performed.

Answer to Problem 28P

The amount of work performed is $\underset{\_}{\text{2}\text{.56}\times {\text{10}}^{-3}\text{J }}$

Explanation of Solution

Write the formula for work in one half revolutions is,     

  $W=U_{\max }-U_{\min }$        (I)

Here, $W$ is the work, $U_{\max }$ is maximum potential energy, and $U_{\min }$ is the minimum potential energy.

Write the formula for potential energy.

  $U=\mu B\cos \theta$        (II)

Here, $B$ is the magnetic field, $\mu$ is the magnetic dipole moment, and $\theta$ is the angle between the $\mu$ and $B$.

Write the formula for torque on a rotor.    

  $\tau =\mu B$        (III)

Here, $\tau$ is the torque,, $B$ is the magnetic field, $\mu$ is the magnetic dipole moment.

Write the formula for torque on a rotor.    

  $\tau =INAB\sin \theta$        (IV)

Here, $\tau$ is the torque, $I$ is current on the particle, $N$ is number of turns, $\theta$ is the angle between $I$ and $B$, and $B$ is the magnetic field.

Equate expression (III) and (IV) and substitute in (V)

Conclusion:

Substitute $180°$ for $\theta$ in $U_{\max }$, $0°$ for $\theta$ in $U_{\min }$  equation (II),

  $\begin{array}{c}W=U_{\max }-U_{\min }\\ =\mu B\cos 180°-\left(-\mu B\cos 0°\right)\\ =2\mu B\end{array}$        (V)

Equate expression (III) and (IV) and substitute in (V),

  $W=2\mu B=2INAB=2\tau$        (VI)

Substitute $\text{6}\text{.40}\times {\text{10}}^{-4}\text{N}\cdot \text{m}$$180°$ for $\tau$ in equation (VI),

    $\begin{array}{c}W=2\left(\text{6}\text{.40}\times {\text{10}}^{-4}\text{N}\cdot \text{m}\right)\\ =\text{1}\text{.28}\times {\text{10}}^{-3}\text{J}\end{array}$

In one full revolution,

    $\begin{array}{c}W=2\left(\text{1}\text{.28}\times {\text{10}}^{-3}\text{J}\right)\\ =\text{2}\text{.56}\times {\text{10}}^{-3}\text{J}\end{array}$

The amount of work performed is $\underset{\_}{\text{2}\text{.56}\times {\text{10}}^{-3}\text{J }}$

(d)

To determine

The average power of the motor.

Answer to Problem 28P

The average power of the motor is $\underset{\_}{\text{0}\text{.154 W}}$

Explanation of Solution

Write the formula for power on a motor.    

  $P_{avg}=\frac{W}{\Delta t}$        (I)

Here, $P_{avg}$ is the power, $W$ is work done, and $\Delta t$ is the time interval.

Conclusion:

The time for revolution is $\Delta t=\frac{\text{60 s}}{\text{3600 rev}}=\frac{1}{60}\text{s}$

Substitute $\text{6}\text{.40}\times {\text{10}}^{-3}\text{N}\cdot \text{m}$  for $W$  , $1/60\text{s}$  for $\Delta t$  equation (I),

  $\begin{array}{c}{P}_{avg}=\frac{\text{6}\text{.40}\times {\text{10}}^{-3}\text{N}\cdot \text{m}}{1/60\text{s}}\\ =\text{0}\text{.154 W}\end{array}$

The average power of the motor is $\underset{\_}{\text{0}\text{.154 W}}$