Electromagnetic Waves

Practice

practice problem 1

Determine the following quanities at a distance of one meter from a 100 W light bulb …
  1. the peak electric field
  2. the peak magnetic field
  3. the radiation pressure

solution

  1. Use the definition of power density and the pointing vector to determine the electric field. Imagine the radiation spreading outward to cover the surface area of a sphere.
     
    S =  P  =  1  EB =  1  E  E  =  E2  ⇒  E = √  Pμ0c
    A μ0 μ0 c μ0c A
     
    Solve for the electric field and imagine the energy spread out over the surface of a 1 m radius sphere.
     
    E = √  Pμ0c  = √  Pμ0c  = √  (100 W)(4π × 10−7 N/A2)(3.00 × 108 m/s)  = √(3000 N2/C2) = 54.8 N/C
    A 4πr2 4π(1 m)2
     
    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).
  2. Use these results and the fact that the ratio of the electric to magnetic fields is the speed of light.
     
    c =  E  ⇒  B =  E  =  54.8 N/C  = 1.83 × 10−7 T = 183 nT
    B c 3.00 × 108 m/s
     
    Compared to the electric field, the magnetic field is less easily detected. For comparison, the earth's magnetic field is 250 times stronger and does not vary much with time.
  3. Use the radiation pressure formula to determine the radiation pressure (duh).
     
    P =  1  ε0E2 =  1  (8.85 × 10−12 C2/Nm2)(54.8 N/C)2 = 1.33 × 10−8 Pa = 13.3 nPa
    2 2
     
    Such a weak pressure is imperceptible under ordinary circumstances and difficult to measure in the laboratory without exceptionally delicate apparatus.

practice problem 2

Write something.

solution

Answer it.

practice problem 3

Write something.

solution

Answer it.

practice problem 4

Write something completely different.

solution

Answer it.

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