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Electric field around a light bulb

  1. May 30, 2010 #1
    1. The problem statement, all variables and given/known data

    What is the peak electric field 2 metres away from a 60W light bulb. Assume that the light emits light evenly in all directions (spherically uniform).

    2. Relevant equations

    I figure that this is to do with the poynting vector.
    [tex]S = \frac{1}{\mu_0}(\vec{E} \times \vec{B})[/tex]

    3. The attempt at a solution

    The total surface integral of the poynting vector at 2 metres away from the bulb should give 60W. From this, the E field can be worked out (Since the magnitude of B is E/mu).

    i.e.

    [tex]\int{\vec{S} \cdot \vec{dA}} = \int{\frac{1}{\mu_0}(\vec{E} \times \vec{B}) \cdot \vec{dA}} [/tex]
    and since the emission is spherically symmetric, S can be taken outside of the integral. This gives that
    [tex]\frac{1}{\mu_0}(\vec{E} \times \vec{B}) \times \frac{4}{3}\pi 2^2 = 60[/tex] (r=2)

    Since [STRIKE][tex]B_0 = \frac{E_0}{\mu_0}[/tex][/STRIKE]

    [tex]E_0 = \frac{B_0}{\mu_0}[/tex]​
    this means that [tex]E_0 = \sqrt{\frac{60 \times {\mu_0}^2}{\frac{4}{3}\pi\times 4}} = 0.0021 Vm^{-1}[/tex]

    Is this correct?
     
    Last edited: May 31, 2010
  2. jcsd
  3. May 30, 2010 #2
    No. The 60 watts isn't radiated away except as blackbody radiation. Although that counts, the exercise might be about the dipole electric field of a 120 volt generator connected to a 240 ohm resistor.

    Edit: you're probably right. The problem says that the light goes in all directions, not that the filament has such and such dimensions.
     
  4. May 31, 2010 #3
    Sorry I noticed an error (area of the sphere)
    [tex]A=4 \pi r^2[/tex]
     
  5. Jun 1, 2010 #4
    Haha, that's embarrassing! I'd also forgotten to square the mu_0 in my numerical answer.

    That gives that the final answer is 0.00000137 Vm^-1 or 1.37 microvolts.
     
  6. Jun 3, 2010 #5
    In addition, B_0 doesn't equal mu_0 times E_0, it equals c times E_0!

    But that gives me an answer of 21 Volts! I'm pretty sure that this would pose some problems in real life, so that's a ridiculous answer! Any help please??
     
  7. Jun 3, 2010 #6

    lanedance

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    Homework Helper

    yeah i think you should use
    B = E/c ;)

    if you write out the equations, its easier to see what you've done
     
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