practice problem 1
Determine the following quanities at a distance of one meter from a 100 W light bulb…
- 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 c = E ⇒ B = E B c P = 1 E E = E2 A μ0 c μ0c 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/mThis 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
- 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
practice problem 2
practice problem 3
practice problem 4
Write something completely different.