Chapter 29, Problem 50P

(a)

To determine

The maximum torque acting on the rotor.

Answer to Problem 50P

The maximum torque acting on the rotor is $6.4\times {10}^{-4}\text{N-m}$.

Explanation of Solution

Write the expression to calculate the maximum torque on the rotor.

    $\tau =nABI$                                                                                                                    (I)

Here, $n$ is the number of turns, $\tau$ is the maximum torque on the rotor, $A$ is the area of the coil, $I$ is the current and $B$ is magnetic field.

Write the expression to calculate the area of the coil.

    $A=lb$                                                                                                                       (II)

Here, $l$ is the length and $b$ is the breadth of the rectangular coil.

Substitute $lb$ for $A$ in equation (I).

    $\tau =n\left(lb\right)BI$                                                                                                            (III)

Conclusion:

Substitute $2.50\text{cm}$ for $l$, $4.00\text{cm}$ for $b$, $80$ for $n$, $0.800\text{T}$ for $B$ and $10.0\text{mA}$ for $I$ in equation (III) to solve for $\tau$.

    $\begin{array}{c}\tau =\left(\begin{array}{l}\left(80\right)\left(2.50\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(4.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(0.800\text{T}\right)\\ \left(10.0\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)\end{array}\right)\\ =6.4\times {10}^{-4}\text{N-m}\end{array}$

Therefore, the charge on the capacitor $C_1$ is $222\mu \text{C}$.

(b)

To determine

The peak power output of the motor.

Answer to Problem 50P

The peak power output of the motor is $0.241\text{W}$.

Explanation of Solution

Write the expression to calculate the peak power output of the  motor.

    $P=\tau \omega$                                                                                                                   (IV)

Here, $P$ is the peak power output of the  motor and $\omega$ is the rotational speed of the rotor.

Conclusion:

Substitute $6.40\times {10}^{-4}\text{N-m}$ for $\tau$ and $3.60\times {10}^3\text{rev}/\text{min}$ for $\omega$ in equation (IV) to solve for $P$.

    $\begin{array}{c}P=\left(6.40\times {10}^{-4}\text{N-m}\right)\left(3.60\times {10}^3\text{rev}/\text{min}\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\min }{60\mathrm{s}}\right)\\ =0.241\text{W}\end{array}$

Therefore, the peak power output of the motor is $0.241\text{W}$.

(c)

To determine

The amount of work performed by the magnetic field on the rotor in every full revolution.

Answer to Problem 50P

The amount of work performed by the magnetic field on the rotor in every full revolution is $4.02\text{mJ}$.

Explanation of Solution

Write the expression to calculate the period of the current.

    $T=\frac{2\pi }{\omega }$                                                                                                                     (V)

Here, $T$ is the period of the current.

Write the expression to calculate the work done.

    $W=PT$                                                                                                                   (VI)

Here, $W$ is the work done by the magnetic field on the rotor in every full revolution.

Substitute $\frac{2\pi }{\omega }$ for $T$ in equation (VI)

    $W=P\left(\frac{2\pi }{\omega }\right)$                                                                                                          (VII)

Conclusion:

Substitute $0.241\text{W}$ for $P$ and $3.60\times {10}^3\text{rev}/\text{min}$ for $\omega$ in equation (VII) to solve for $W$.

    $\begin{array}{c}W=\left(0.241\text{W}\right)\left(\frac{2\pi }{3.60\times {10}^3\text{rev}/\text{min}\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\min }{60\mathrm{s}}}\right)\\ =4.02\times {10}^{-3}\text{J}\times \frac{{10}^3\text{mJ}}{\text{J}}\\ =4.02\text{mJ}\end{array}$

Therefore, the amount of work performed by the magnetic field on the rotor in every full revolution is $4.02\text{mJ}$.

(d)

To determine

The average power of the motor.

Answer to Problem 50P

The average power of the motor is $0.17\text{W}$.

Explanation of Solution

Write the expression to calculate the average power.

    $P_{\text{avg}}=\frac{P}{\sqrt{2}}$                                                                                                             (VIII)

Here, $P_{\text{avg}}$ is the average power.

Conclusion:

Substitute $0.241\text{W}$ for $P$ in equation (VIII) to solve for $P_{\text{avg}}$.

    $\begin{array}{c}{P}_{\text{avg}}=\frac{0.241\text{W}}{\sqrt{2}}\\ =0.17\text{W}\end{array}$

Therefore, the average power of the motor is $0.17\text{W}$.