Chapter 30, Problem 48CP

(a)

To determine

The average power delivered to the disk.

Answer to Problem 48CP

The average power delivered to the disk is $\frac{\pi R^4{\omega }^2{B}^2{}_{\max }b}{16\rho }$.

Explanation of Solution

Given info: Radius of disk is $R$, thickness of disk is $b$, resistivity of the disk is $\rho$ and sinusoidal magnetic field is $B_{\max }\cos \omega t$.

Since the eddy currents occur as concentric circles with the disk. Consider the disk to be a collection of rings that each has an induced emf.

The emf induced in the disk can be given as,

$\epsilon =-\frac{d\left(BA\right)}{dt}$

Here,

$\epsilon$ is the emf induced in the disk.

$B$ is the magnetic field.

$A$ is the area of the disk.

Substitute $B_{\max }\cos \omega t$ for $B$ and $\pi r^2$ for $A$ in the above equation,

$\begin{array}{c}\epsilon =-\frac{d\left(B_{\max }\cos \omega t\right)\left(\pi r^2\right)}{dt}\\ =\pi r^2\omega B_{\max }\sin \omega t\end{array}$

Here,

$B_{\max }$ is the maximum magnetic field.

$\omega$ is the angular velocity.

$r$ is the radius of the disk.

$t$ is the time period of the disk.

The elemental resistance around the ring can be given as,

$dR=\frac{\rho l_{\text{ring}}}{A_{\text{ring}}}$

Here,

$R$ is resistance in the ring

$\rho$ is the resistivity of the disk

$l_{\text{ring}}$ is the length of the elemental ring

$A_{\text{ring}}$ is the area of the elemental ring

Substitute $2\pi r$ for $l_{\text{ring}}$ and $bdr$ for $A_{\text{ring}}$ in the above equation,

$dR=\frac{\rho \left(2\pi r\right)}{bdr}$

The power delivered to the elemental ring can be given as,

$dP=\frac{{\epsilon }^2}{dR}$

$P$ is the power delivered to the ring,

Substitute $\frac{\rho \left(2\pi r\right)}{bdr}$ for $dR$ and $\pi r^2\omega B_{\max }\sin \omega t$ for $\epsilon$ in the above equation,

$\begin{array}{c}dP=\frac{{\left(\pi r^2\omega B_{\max }\sin \omega t\right)}^2}{\left(\frac{\rho \left(2\pi r\right)}{bdr}\right)}\\ =\frac{\pi r^3{\omega }^2{B}^2{}_{\max }{\sin }^2\left(\omega t\right)bdr}{2\rho }\end{array}$

The total power delivered to the disk can be given as,

$P={\displaystyle \int dP}$

Substitute $\frac{\pi r^3{\omega }^2{B}^2{}_{\max }{\sin }^2\left(\omega t\right)bdr}{2\rho }$ for $dP$ in the above equation,

$\begin{array}{c}P={\displaystyle \underset{0}{\overset{R}{\int }}\frac{\pi r^3{\omega }^2{B}^2{}_{\max }{\sin }^2\left(\omega t\right)bdr}{2\rho }}\\ =\frac{\pi {\omega }^2{B}^2{}_{\max }b{\sin }^2\left(\omega t\right)}{2\rho }{\displaystyle \underset{0}{\overset{R}{\int }}{r}^{3}dr}\\ =\frac{\pi R^4{\omega }^2{B}^2{}_{\max }{\sin }^2\left(\omega t\right)bdr}{8\rho }\end{array}$

Substitute $\frac{1}{2}$ for ${\sin }^2\left(\omega t\right)$ to get average power in the above equation,

$P_{\text{avg}}=\frac{\pi R^4{\omega }^2{B}^2{}_{\max }b}{16\rho }$ (1)

Here,

$P_{\text{avg}}$ is the average power delivered to the disk.

Thus, the average power delivered to the disk can be given as $\frac{\pi R^4{\omega }^2{B}^2{}_{\max }b}{16\rho }$.

Conclusion:

Therefore, the average power delivered to the disk can be given as $\frac{\pi R^4{\omega }^2{B}^2{}_{\max }b}{16\rho }$.

(b)

To determine

The factor by which power will change when the field doubles.

Answer to Problem 48CP

The factor by which power will change when the field doubles is four times.

Explanation of Solution

Given info: Radius of disk is $R$, thickness of disk is $b$, resistivity of the disk is $\rho$ and sinusoidal magnetic field is $B_{\max }\cos \omega t$.

The relation between the field and the power can be given from equation (1) as,

$P_{\text{avg}}\propto B^2{}_{\max }$

Substitute $2B_{\max }$ for $B_{\max }$ in the above equation,

$\begin{array}{c}{P}_{\text{avg}}\propto {\left(2B\right)}^2{}_{\max }\\ \propto 4B^2{}_{\max }\\ \propto \text{4}{P}_{\text{avg}}\end{array}$

Thus, the power will change by four times when the field doubles.

Conclusion:

Therefore, the factor by which power will change when the field doubles is four times.

(c)

To determine

The factor by which power will change when the frequency doubles.

Answer to Problem 48CP

The factor by which power will change when the frequency doubles is four times.

Explanation of Solution

Given info: Radius of disk is $R$, thickness of disk is $b$, resistivity of the disk is $\rho$ and sinusoidal magnetic field is $B_{\max }\cos \omega t$.

The relation between the field and the power can be given from equation (1) as,

$P_{\text{avg}}\propto {\omega }^2$

Substitute $2\pi f$ for $\omega$ in the above equation,

$\begin{array}{c}{P}_{\text{avg}}\propto {\left(2\pi f\right)}^2\\ \propto f^2\end{array}$

Here,

$f$ is the frequency of the disk.

Substitute $2f$ for $f$ in the above equation,

$\begin{array}{c}{P}_{\text{avg}}\propto {\left(2f\right)}^2\\ \propto 4f^2\\ \propto \text{4}{P}_{\text{avg}}\end{array}$

Thus, the power will change by four times when the frequency doubles.

Conclusion:

Therefore, the factor by which power will change when the frequency doubles is four times.

(d)

To determine

The factor by which power will change when the radius of the disk doubles.

Answer to Problem 48CP

The factor by which power will change when the radius of the disk doubles is sixteen times.

Explanation of Solution

Given info: Radius of disk is $R$, thickness of disk is $b$, resistivity of the disk is $\rho$ and sinusoidal magnetic field is $B_{\max }\cos \omega t$.

The relation between the field and the power can be given from equation (1) as,

$P_{\text{avg}}\propto R^4$

Substitute $2R$ for $R$ in the above equation,

$\begin{array}{c}{P}_{\text{avg}}\propto {\left(2R\right)}^4\\ \propto 16R^4\\ \propto 16P_{\text{avg}}\end{array}$

Thus, the power will change by sixteen times when the radius of disk doubles.

Conclusion:

Therefore, the factor by which power will change when the radius of disk doubles is sixteen times.