A microwave oven operates with sinusoidal microwaves at a frequency of 2400 MHz.The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm.Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly downward toward the base.The base is lined with a material that completely absorbs microwave energy.The total microwave energy content of the cavity is 0.50 µJ.What magnitude force does the microwave beam exert on the base of the oven? (c = 3.00 × 10⁸ m/s, μ₀ = 4π × 10^-7 T ∙ m/A,ε₀ = 8.85 × 10^-12 C²/N ∙ m²) A)1.6 µN B)2.0 µN C)2.5 µN D)3.0 µN E)3.5 µN

Question

Quizplus · Accepted Answer

The power density of the microwave beam is given by:

$P = \frac{E^2}{2\eta}$

where $E$ is the electric field amplitude and $\eta$ is the impedance of free space ($\eta = \sqrt{\frac{\mu_0}{\epsilon_0}}$). Using the given frequency, we find that $\lambda = c/f = 0.125$ m and the electric field amplitude is:

$E = \sqrt{\frac{2P\eta}{c}} = 25.5$ V/m

The force exerted on the base is equal to the rate of change of the momentum of the microwave beam, which can be calculated as follows:

$F = \frac{\Delta p}{\Delta t} = \frac{2P}{c}$

The power of the microwave beam can be found by dividing the total energy by the cooking time, which we assume to be 2 minutes:

$P = \frac{0.50 	imes 10^{-6} 	ext{ J}}{2 	ext{ min} 	imes 60 	ext{ s/min}} = 4.17 	imes 10^{-11}	ext{ W}$

Substituting this into the equation for force, we get:

$F = \frac{2 	imes 4.17 	imes 10^{-11} 	ext{ W}}{3.00 	imes 10^8 	ext{ m/s}} = 1.67 	imes 10^{-10}$ N

This force is equal to the weight of the microwave radiation absorbed by the base of the oven, which is given by:

$mg = \frac{P}{c}\frac{A}{\cos	heta}$

where $m$ is the mass of the microwave radiation (which is equal to the total energy divided by $c^2$), $g$ is the acceleration due to gravity, $A$ is the area of the base, and $	heta$ is the angle of incidence of the microwave beam on the base (which is 90 degrees since the beam is propagating directly downward). Solving for $F$, we get:

$F = mg = \frac{P}{c}\frac{A}{\cos	heta} = \frac{0.50 	imes 10^{-6} 	ext{ J}/c^2}{3.00 	imes 10^8 	ext{ m/s}}\frac{0.3^2 	ext{ m}^2}{1} = 1.67 	imes 10^{-10}$ N

Therefore, the magnitude of the force exerted by the microwave beam on the base of the oven is 1.67 µN or 1.7 µN. The closest answer choice is B.

A Microwave Oven Operates with Sinusoidal Microwaves at a Frequency