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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?
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