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

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Homework Help Overview

The discussion revolves around calculating the peak electric field at a distance of 2 meters from a 60W light bulb, assuming isotropic emission of light. The context involves concepts from electromagnetism, particularly the Poynting vector and electric and magnetic field relationships.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the Poynting vector to relate power and electric field strength. There are attempts to derive the electric field from the power output and surface area of a sphere. Some participants question the assumptions regarding the nature of the light bulb's emission and the relevance of blackbody radiation.

Discussion Status

There are multiple interpretations of the problem, with participants exploring different aspects of the electric and magnetic fields. Some have pointed out errors in calculations and assumptions, while others have suggested alternative approaches to the problem. Guidance has been offered regarding the relationships between electric and magnetic fields.

Contextual Notes

Participants note potential errors in the area calculation for the sphere and the relationships between the electric and magnetic fields, indicating a need for careful consideration of the physics involved. There is also mention of practical implications of the calculated values, suggesting a concern for realism in the results.

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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?
 
Last edited:
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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.
 
Sorry I noticed an error (area of the sphere)
A=4 \pi r^2
 
zzzoak said:
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.
 
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??
 
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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