Question

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 is the amplitude of the electric field? (c = 3.00 × 108 m/s, μ0 = 4π × 10-7 T ∙ m/A, ε0 = 8.85 × 10-12 C2/N ∙ m2) Group of answer choices

Answer 1

Answer:

1584.6 V/m

Explanation:

The energy density (the energy per unit volume) for an electromagnetic wave is given by the equation

$w=\epsilon_0 E^2$

where

$\epsilon_0 = 8.85\cdot 10^{-12} C^2/Nm^2$ is the vacuum permittivity

E is the amplitude of the electric field of the wave

At the same time, the energy density can be written as

$w=\frac{U}{V}$

where

U is the total energy of the wave

V is the volume considered

So the first equation can be written as

$U=\epsilon_0 V E^2$

Or

$E=\sqrt{\frac{U}{\epsilon_0 V}}$

In this problem:

$U=0.50 \mu J = 0.50\cdot 10^{-6} J$ is the total energy of the microwave

The volume considered is the volume of the cavity, so:

$V=0.25 m \cdot 0.30 m \cdot 0.30 m = 0.0225 m^3$

Therefore, the amplitude of the electric field is:

$E=\sqrt{\frac{0.50\cdot 10^{-6}}{(8.85\cdot 10^{-12})(0.0225)}}=1584.6 V/m$