practice problem 1
- the peak electric field
- the peak magnetic field
- the radiation pressure
- Set the definition of power density equal to the poynting vector.
power density P = 1 EB poynting vector A μ0
The speed of light is the ratio of the electric to magnetic fields. Rearrange this equation.
c = E ⇒ B = E B c
Substitute the second equation into the first to eliminate the magnetic field.
P = 1 E E = E2 A μ0 c μ0c
Solve for the electric field and imagine the energy spread out over the surface of a 1 m radius sphere.
E = √ Pμ0c = √ Pμ0c A 4πr2 E = √ (100 W)(4π × 10−7 N/A2)(3.00 × 108 m/s) 4π(1 m)2
E = √(3000 N2/C2) = 54.8 N/C or 54.8 V/m
This is a field strength that could be measured with fairly inexpensive equipment if it weren't fluctuating so rapidly. (Visible light frequencies are very high). For comparison, the average fair weather electric field on the surface of the Earth (120 V/m) is only about twice as big.
- Substitute back into the second equation to determine the magnetic field.
B = E = 54.8 N/C = 1.83 × 10−7 T = 183 nT c 3.00 × 108 m/s
Again, this is a field strength that could be measured with fairly inexpensive equipment if it weren't fluctuating so rapidly. For comparison, the average magnetic field on the surface of the Earth (45 µT) is 250 times stronger and does not vary much with time.
- Use the radiation pressure equation to determine the radiation pressure (duh).
P = 1 ε0E2 2 P = 1 (8.85 × 10−12 C2/Nm2)(54.8 N/C)2 2 P = 1.33 × 10−8 Pa = 13.3 nPa
Such a weak pressure is imperceptible under ordinary circumstances and difficult to measure in the laboratory without delicate apparatus. For comparison, standard atmospheric pressure (101 kPa) is 8 trillion times stronger. Light pressure is easy to ignore.
practice problem 2
practice problem 3
practice problem 4