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Chapter 29: Problem 45

High microwave intensities can cause biological damage through heating of tissue; a particular concern is cataract formation. The U.S. Food and Drug Administration limits microwave radiation near the door of a microwave oven to \(5.0 \mathrm{mW} / \mathrm{m}^{2} .\) The window in a particular oven door measures \(40 \mathrm{cm}\) by \(17 \mathrm{cm}\) and is covered with a metal screen to block microwaves. Assuming power leaks uniformly through the window area, what percent of the oven's \(900-\mathrm{W}\) microwave power can leak without exceeding the FDA standards?

Short Answer

Expert verified
The oven's microwave power can leak up to approximately 0.0378% without exceeding the FDA standard.

Step by step solution

01

Convert quantities to the same units

To make calculations easier, convert all units to the standard international (SI) units. The area of the window is given in cm but should be converted to m. So the dimensions become \(0.4 \, \mathrm{m}\) by \(0.17 \, \mathrm{m}\).
02

Calculate window area in square meters

We calculate the area of the window using the formula for the area of a rectangle, which is length times width. So the area \(A\) is \(0.4 \, \mathrm{m} \times 0.17 \, \mathrm{m} = 0.068 \, \mathrm{m}^{2}\).
03

Calculate allowable leakage power

The intensity \(I\) of the microwave radiation is the power \(P\) per unit area \(A\). Therefore, we can find the allowable power leakage by multiplying the intensity limit by the area. So the allowed power \(P_{allowed}\) is \(5.0 \, \mathrm{mW} / \mathrm{m}^{2} \times 0.068 = 0.34 \, \mathrm{W}\).
04

Calculate the percentage of power that can leak

Finally, we find the percentage of the oven's total microwave power that the leakage represents by dividing the allowable leakage power by the total power, and multiply by 100 to get the percentage. So the allowable leakage percentage is \((0.34 \, \mathrm{W} / 900 \, \mathrm{W}) \times 100 = 0.0378 \% \).

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