Chapter 21, Problem 76AP

(a)

To determine

The radiation intensity of a cell phone at a distance of $2.0\text{inch}$.

Answer to Problem 76AP

The radiation intensity of a cell phone at a distance of $2.0\text{inch}$ is $6.2\text{mW}/{\text{cm}}^2$.

Explanation of Solution

Given Info: Distance $r$ is $2.0\text{inch}$, Power of radiation emitted from the cell phoneis $2.0\text{W}$.

Formula to calculate the radiation intensity $I$ from a cell phone is,

$I=\frac{P}{A}$

• I is the radiation intensity
• P is the output peak power of the cell phone

Formula to calculate the area of the sphere is,

$A=4\pi r^2$

• A is the area of the sphere
•  r is the distance

Use $4\pi r^2$ for $A$ in the above equation to find the expression for $I$.

$I=\frac{P}{4\pi r^2}$

Substitute $2.0\text{in}$ for $r$, $2.0\text{W}$ for P in the above equation to find $I$.

$\begin{array}{c}I=\frac{\left(2.0\text{W}\right)\left(\frac{1\text{mW}}{{10}^{-3}\text{W}}\right)}{4\pi {\left[\left(2.0\text{in}\right)\left(\frac{2.54\text{cm}}{1\text{inch}}\right)\right]}^2}\\ =\frac{\left(2.0\times {10}^3\text{mW}\right)}{4\pi {\left(5.1\text{cm}\right)}^2}\\ =61.22\text{mW}/{\text{cm}}^2\\ \approx 6.2\text{mW}/{\text{cm}}^2\end{array}$

The radiation intensity of a cell phone at a distance of $2.0\text{inch}$ is $6.2\text{mW}/{\text{cm}}^2$.

Conclusion:

Formula to calculate the percentage increase is,

$\text{Percentage increase}=\frac{\left(I-I_0\right)}{I}\times 100$

• $I_0$ is the maximum allowable microwave leakage

Substitute $6.2\text{mW}/{\text{cm}}^2$ for $I$ and $5.0\text{mW}/{\text{cm}}^2$ for $I_0$ to find $\text{Percentage increase}$ of radiation.

$\begin{array}{c}\%\text{higher}=\frac{\left(6.2\text{mW}/{\text{cm}}^2-5\text{mW}/{\text{cm}}^2\right)}{6.2\text{mW}/{\text{cm}}^2}\times 100\%\\ =24\%\end{array}$

The intensity $6.2\text{mW}/{\text{cm}}^2$ is $24\text{\%}\text{higher}$ that of the maximum allowed leakage of $5.0\text{mW}/{\text{cm}}^2$ from a microwave at the distance of $2.0\text{inches}$.

(b)

To determine

The radiation intensity $I_{\text{total}}$ in the brain if the Bluetooth headset is connected.

Answer to Problem 76AP

The radiation intensity $I_{\text{total}}$ in the brain if the Bluetooth headset is connected is $0.024\text{mW}/{\text{cm}}^2$.

Explanation of Solution

Section 1:

To determine: The intensity $I_{\text{phone}}$ of radiation from a cell phone.

Answer: The intensity $I_{\text{phone}}$ of radiation from a cell phone is $1.6\times {10}^{-2}\text{mW}/{\text{cm}}^2$.

Explanation:

Given Info: Radiation power $P_{\text{c}}$ is $2.0\text{W}$ and Distance between the headset and the cell phone in the pocket $r_{\text{p}}$ is $1.0\text{m}$.

Formula to calculate the radiation intensity $I_{\text{phone}}$ of a cell phone is,

$I_{\text{phone}}=\frac{P_{\text{c}}}{A_{\text{P}}}$

• $P_{\text{c}}$ is the output peak power of the cell phone
• $A_{\text{P}}$ is the area of intensity sphere between the cell phone and headset.
• $I_{\text{phone}}$ is the intensity of radiation a cell phone

Use $4\pi r_{\text{p}}{}^2$ for $A_{\text{P}}$ in the above equation to find the expression for $I_{\text{phone}}$.

$I_{\text{phone}}=\frac{P_{\text{c}}}{4\pi r_{\text{p}}{}^2}$

• $r_{\text{p}}$ is the distance between the headset and the cell phone in the pocket.

Substitute $1.0\text{m}$ for $r_{\text{p}}$, $2.0\text{W}$ for $P_{\text{c}}$ in the above equation to find $I_{\text{phone}}$.

$\begin{array}{c}{I}_{\text{phone}}=\frac{\left(2.0\text{W}\right)\left(\frac{1\text{mW}}{{10}^{-3}\text{W}}\right)}{4\pi \left(1.0\text{m}\right){\left(\frac{1\text{cm}}{{10}^{-2}\text{m}}\right)}^2}\\ =\frac{\left(2.0\times {10}^3\text{mW}\right)}{4\pi {\left(1.0\times {10}^2\text{cm}\right)}^2}\\ =1.6\times {10}^{-2}\text{mW}/{\text{cm}}^2\end{array}$

Thus the radiation intensity $I_{\text{phone}}$ of a cell phone is $1.6\times {10}^{-2}\text{mW}/{\text{cm}}^2$

Section 2:

To determine: The radiation intensity $I_{\text{headset}}$ of a headset.

Answer: The radiation intensity $I_{\text{headset}}$ of a cell phone is $7.6\times {10}^{-3}\text{mW}/{\text{cm}}^2$

Explanation:

Given Info: Distance between the Bluetooth headset and the brain $r_{\text{h}}$ is $2.0\text{inch}$, Bluetooth headset radiating power $P_{\text{h}}$ is $2.5\text{mW}$.

Formula to calculate the radiation intensity $I_{\text{headset}}$ of a cell phone is,

$I_{\text{headset}}=\frac{P_{\text{h}}}{A_{\text{h}}}$

• $I_{\text{headset}}$ is the radiation intensity of a headset,
• $P_{\text{h}}$ is the output peak power of the headset,
• $A_{\text{h}}$ is the area of intensity sphere,

Use $4\pi r_{\text{h}}{}^2$ for $A_{\text{h}}$ in the above equation to find the expression for $I_{\text{headset}}$.

$I=\frac{P_{\text{h}}}{4\pi r_{\text{h}}{}^2}$

• $r_{\text{h}}$ is the distance between the headset and the cell phone in the pocket.

Substitute $2.0\text{inch}$ for $r_{\text{h}}$, $2.5\text{W}$ for $P_{\text{h}}$ in the above equation to find $I_{\text{headset}}$.

$\begin{array}{c}{I}_{\text{headset}}=\frac{\left(2.5\text{W}\right)\left(\frac{1\text{mW}}{{10}^{-3}\text{W}}\right)}{4\pi {\left[\left(2.0\text{inch}\right)\left(\frac{2.54\text{cm}}{1\text{inch}}\right)\right]}^2}\\ =\frac{\left(2.5\times {10}^3\text{mW}\right)}{4\pi {\left(5.1\text{cm}\right)}^2}\\ =7.6\times {10}^{-3}\text{mW}/{\text{cm}}^2\end{array}$

Thus the radiation intensity $I_{\text{headset}}$ of a cell phone is $7.6\times {10}^{-3}\text{mW}/{\text{cm}}^2$

Conclusion:

Total intensity of radiation is the sum of radiation emitted by the cell phone and the radiation emitted by the headset.

Formula to calculate the radiation intensity $I_{\text{total}}$ is,

$I_{\text{total}}=I_{\text{phone}}+I_{\text{headset}}$

• $I_{\text{total}}$ is the total radiation

Substitute $1.6\times {10}^{-2}\text{mW}/{\text{cm}}^2$ for $I_{\text{phone}}$ and $7.6\times {10}^{-3}\text{mW}/{\text{cm}}^2$ for $I_{\text{headset}}$ in the above equation to find $I_{\text{total}}$.

$\begin{array}{c}{I}_{\text{total}}=\left(1.6\times {10}^{-2}\frac{\text{mW}}{{\text{cm}}^2}\right)+\left(7.6\times {10}^{-3}\frac{\text{mW}}{{\text{cm}}^2}\right)\\ =0.024\frac{\text{mW}}{{\text{cm}}^2}\end{array}$

Therefore the radiation intensity $I_{\text{total}}$ in the brain if the Bluetooth headset is connected is $0.024\text{mW}/{\text{cm}}^2$