Chapter 22, Problem 84P

(a)

To determine

The intensity of the light.

Answer to Problem 84P

The intensity of the light is $170{\text{W/m}}^2$.

Explanation of Solution

Write the equation to find the power of the lamp.

    $P=I_{rms}\Delta V_{rms}$                                                                                           (I)

Here, $P$ is the power of the lamp, $I_{rms}$ is the rms current and $\Delta V_{rms}$ is the rms voltage.

Find the equation for the intensity and substitute $2\pi r^2$ for $A$.

    $\begin{array}{c}I=\frac{P}{A}\\ =\frac{P}{2\pi r^2}\end{array}$                                                                                                     (II)

Here, $I$ is the intensity and $r$ is the radius.

Conclusion:

Substitute $12.5\text{A}$ for $I_{rms}$ and $120\text{V}$ for $\Delta V_{rms}$ in equation (I) to find $P$.

    $\begin{array}{c}P=\left(12.5\text{A}\right)\left(120\text{V}\right)\\ =1500\text{W}\end{array}$

Substitute $1500\text{W}$ for $P$ and $1.2\text{m}$ for $r$ in equation (II) to find $\lambda$.

    $\begin{array}{c}I=\frac{1500\text{W}}{2\pi {\left(1.2\text{m}\right)}^2}\\ =\frac{1500\text{W}}{2\pi \left(1.44\text{m}\right)}\\ =170{\text{W/m}}^2\end{array}$

Thus, the intensity of the light is $170{\text{W/m}}^2$.

(b)

To determine

The energy incident on the face.

Answer to Problem 84P

The energy incident on the face is $560\text{J}$.

Explanation of Solution

Write the equation connecting power, energy and intensity.

    $P_{face}=\frac{\Delta U_{face}}{\Delta t}=I{A}_{face}$                                                                                       (III)

Here, $P_{face}$ is the power of the light incident on the face, $\Delta U_{face}$ is the energy energy incident on the face, $\Delta t$ is the time and $A_{face}$ is the area of the face.

Rearrange and substitute $\frac{P}{2\pi r^2}$ for $I$ to find $\Delta U_{face}$.

    $\begin{array}{c}\Delta U_{face}=I{A}_{face}\Delta t\\ =\frac{P}{2\pi r^2}{A}_{face}\Delta t\end{array}$                                                                                      (IV)

Conclusion:

Substitute $1500\text{W}$ for $P$, $2.8\times {10}^{-2}{\text{m}}^2$ for $A_{face}$, $2.0\text{min}$ for $\Delta t$ and $1.2\text{m}$ for $r$ in equation (IV) to find $\Delta U_{face}$.

    $\begin{array}{c}\Delta U_{face}=\frac{1500\text{W}}{2\pi {\left(1.2\text{m}\right)}^2}\left(2.8\times {10}^{-2}{\text{m}}^2\right)\left(2.0\text{min}\right)\\ =\frac{1500\text{W}}{2\pi {\left(1.2\text{m}\right)}^2}\left(2.8\times {10}^{-2}{\text{m}}^2\right)\left(2.0\text{min}\right)\left(\frac{60\text{s}}{1\text{min}}\right)\\ =560\text{J}\end{array}$

Thus, the energy incident on the face is $560\text{J}$.

(c)

To determine

The rms electric and magnetic fields.

Answer to Problem 84P

The rms electric field is $250\text{V/m}$ and the rms magnetic field is $8.3\times {10}^{-7}\text{T}$.

Explanation of Solution

Write the equation for the rms electric field substituting $\frac{P}{2\pi r^2}$ for $I$.

    $\begin{array}{c}{E}_{rms}=\sqrt{\frac{I}{{\epsilon }_{0}c}}\\ =\sqrt{\frac{P}{2\pi r^2}\frac{1}{{\epsilon }_{0}c}}\end{array}$                                                                                           (V)

Here, $E_{rms}$ is the rms electric field, ${\epsilon }_0$ is the absolute permittivity and $c$ is the speed of light.

Write the equation for the rms magnetic field.

    $B_{rms}=\frac{E_{rms}}{c}$                                                                                          (VI)

Here, $B_{rms}$ is the rms magnetic field.

Conclusion:

Substitute $1500\text{W}$ for $P$, $3.00\times {10}^8\text{m/s}$ for $c$, $8.854\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$ and $1.2\text{m}$ for $r$ in equation (V) to find $E_{rms}$.

    $\begin{array}{c}{E}_{rms}=\sqrt{\frac{1500\text{W}}{2\pi {\left(1.2\text{m}\right)}^2}\frac{1}{\left(8.854\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}\right)\left(3.00\times {10}^8\text{m/s}\right)}}\\ =\sqrt{\frac{1500\text{W}}{2\pi \left(1.44\text{m}\right)}\frac{1}{\left(8.854\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}\cdot {\text{m}}^{\text{2}}\right)\left(3.00\times {10}^8\text{m/s}\right)}}\\ =250\text{V/m}\end{array}$

Substitute $250\text{V/m}$ for $E_{rms}$ and $3.00\times {10}^8\text{m/s}$ for $c$ in equation (VI) to find $B_{rms}$.

    $\begin{array}{c}{B}_{rms}=\frac{250\text{V/m}}{3.00\times {10}^8\text{m/s}}\\ =8.3\times {10}^{-7}\text{T}\end{array}$

Thus, the rms electric field is $250\text{V/m}$ and the rms magnetic field is $8.3\times {10}^{-7}\text{T}$.