What is the Peak Electric Field 2 Metres from a 60W Light Bulb?

[DeShark]

Homework Statement

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).

Homework Equations

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

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.

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

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

$$E_0 = \frac{B_0}{\mu_0}$$​

this means that $$E_0 = \sqrt{\frac{60 \times {\mu_0}^2}{\frac{4}{3}\pi\times 4}} = 0.0021 Vm^{-1}$$

Is this correct?

 

[Antiphon]

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.

 

[zzzoak]

Sorry I noticed an error (area of the sphere)
$$A=4 \pi r^2$$

 

[DeShark]

Sorry I noticed an error (area of the sphere)
$$A=4 \pi r^2$$

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.

 

[DeShark]

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??

 

[lanedance]

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

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