Chapter 29, Problem 29.50P

(a)

To determine

The maximum torque acting on the rotor.

Answer to Problem 29.50P

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

Explanation of Solution

Given info: The number of turns in the rectangular coil is $80$, magnetic field is $0.800\text{T}$, electric current is $10.00\text{mA}$, length of the coil is $2.50\text{cm}$ and breadth of the coil is $4.00\text{cm}$.

The formula to calculate the area of the coil is,

$A=ab$

Here,

$a$ is the length of the coil.

$b$ is the breadth of the coil.

Substitute $2.50\text{cm}$ for $a$ and $4.00\text{cm}$ for $b$ in the above formula to find $A$.

$\begin{array}{l}A=ab\\ =\left(2.50\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(4.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =10.0\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

Thus, the area of the coil is $10.0\times {10}^{-4}{\text{m}}^{\text{2}}$.

The formula to calculate the torque is,

$\tau =nABI$

Here,

$n$ is the number of turns of the rectangular coil.

$B$ is the magnetic field.

$A$ is the area of the coil.

$I$ is the electric current.

Substitute $10.00\text{mA}$ for $I$ and $10.0\times {10}^{-4}{\text{m}}^{\text{2}}$ for $A$ , $0.800\text{T}$ for $B$, $80$ for $n$ in the above formula to find $\mu$.

$\begin{array}{c}\tau =nABI\\ =80\left(10.00\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)\left(10.0\times {10}^{-4}{\text{m}}^{\text{2}}\right)\left(0.800\text{T}\right)\\ =6.4\times {10}^{-4}\text{N}\cdot \text{m}\end{array}$

Conclusion:

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

(b)

To determine

The peak power output of the motor.

Answer to Problem 29.50P

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

Explanation of Solution

Given info: The number of turns in the rectangular coil is $80$, magnetic field is $0.800\text{T}$, electric current is $10.00\text{mA}$, length of the coil is $2.50\text{cm}$ and breadth of the coil is $4.00\text{cm}$.

The formula to calculate the peak power is,

$P=\tau \omega$

Here,

$\tau$ is the torque.

$\omega$ is the angular frequency.

Substitute $6.4\times {10}^{-4}\text{N}\cdot \text{m}$ for $\tau$ and $3.60\times {10}^3\text{rev}/\text{min}$ in the above formula to find $P$.

$\begin{array}{c}P=\tau \omega \\ =\left(6.4\times {10}^{-4}\text{N}\cdot \text{m}\right)\left(3.60\times {10}^3\text{rev}/\text{min}\right)\left(\frac{2\pi \text{rad}/\text{s}}{60\text{rev}/\text{min}}\right)\\ =0.24115\text{W}\\ =\text{0}\text{.241}\text{W}\end{array}$

Conclusion:

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 29.50P

The amount of work performed by the magnetic field on the rotor in every full revolution is $2.56\text{mJ}$.

Explanation of Solution

Given info: The number of turns in the rectangular coil is $80$, magnetic field is $0.800\text{T}$, electric current is $10.00\text{mA}$, length of the coil is $2.50\text{cm}$ and breadth of the coil is $4.00\text{cm}$.

The formula to calculate the work done in half a revolution is,

$W=n\mu B\left(\cos {\theta }_1-\cos {\theta }_2\right)$

Here,

$n$ is the number of turns in the coil.

$\mu$ is the magnetic moment.

$B$ is the magnetic field.

${\theta }_1$ is the initial angle between the magnetic field and magnetic moment.

${\theta }_2$ is the final angle between the magnetic field and magnetic moment.

The formula to calculate the magnetic moment is,

$\mu =IA$

Here,

$I$ is the electric current.

$A$ is the area of the coil.

Substitute $10.00\text{mA}$ for $I$ and $10.0\times {10}^{-4}{\text{m}}^{\text{2}}$ for $A$ in the above formula to find $\mu$.

$\begin{array}{c}\mu =IA\\ =\left(10.00\text{mA}\right)\left(10.0\times {10}^{-4}{\text{m}}^{\text{2}}\right)\\ =0.01\text{A}\cdot {\text{m}}^{\text{2}}\end{array}$

Thus, the magnetic moment is $0.01\text{A}\cdot {\text{m}}^{\text{2}}$.

Substitute $0.01\text{A}\cdot {\text{m}}^{\text{2}}$ for $\mu$, $80$ for $n$ , $10.00\text{mA}$ for $I$, $0.800\text{T}$ for $B$ $0°$ for ${\theta }_1$ and $180°$ for ${\theta }_2$  in the above formula to find $W$.

$\begin{array}{c}W=n\mu B\left(\cos {\theta }_1-\cos {\theta }_2\right)\\ =80\left(0.01\text{A}\cdot {\text{m}}^{\text{2}}\right)\left(0.800\text{T}\right)\left(\cos 0°-\cos 180°\right)\\ =80\left(0.01\text{A}\cdot {\text{m}}^{\text{2}}\right)\left(0.800\text{T}\right)\left(1-\left(-1\right)\right)\\ =1.28\text{mJ}\end{array}$

Thus, the work done in half the revolution is $1.26\text{mJ}$.

The formula to calculate the work done to complete a full revolution is,

$W^{\prime }=2W$

Here,

$W$ is the work done to complete half a revolution.

Substitute $1.28\text{mJ}$ for $W$ in the above formula to find $W^{\prime }$.

$\begin{array}{c}{W}^{\prime }=2W\\ =2\left(1.28\text{}\text{mJ}\right)\\ =2.56\text{mJ}\end{array}$

Conclusion:

Therefore, the amount of work performed by the magnetic field on the rotor in every full revolution is $2.56\text{mJ}$.

(d)

To determine

The average power of the motor.

Answer to Problem 29.50P

The average power of the motor is $0.154\text{W}$.

Explanation of Solution

Given info: The number of turns in the rectangular coil is $80$, magnetic field is $0.800\text{T}$, electric current is $10.00\text{mA}$, length of the coil is $2.50\text{cm}$ and breadth of the coil is $4.00\text{cm}$.

The formula to calculate the average output power is,

$P_{\text{avg}}=\frac{W}{t}$

Here,

$W$ is the work done to perform one complete revolution.

$t$ is the time taken to complete one revolution.

Substitute $2.56\text{J}$ for $W$ and $1\text{rev}/\text{min}$ for $t$ in the above formula to find $P_{\text{avg}}$.

$\begin{array}{c}{P}_{\text{avg}}=\frac{W}{t}\\ =\frac{2.56\text{mJ}\times \frac{{10}^{-3}\text{J}}{1\text{mJ}}}{1\text{rev}/\text{min}\times \frac{\frac{1}{60}\text{rev}/\text{s}}{1\text{rev}/\text{min}}}\\ =0.154\text{W}\end{array}$

Conclusion:

Therefore, the average power of the motor is $0.154\text{W}$.