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Fields from a Light Bulb. We can reasonably model a 75 - W incandescent light bulb as a sphere 6.0 cm in diameter. Typically, only about 5% of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible-light intensity (in ) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?

Short Answer

Expert verified

a. The intensity of visible-light at the surface of the bulb 330W/m2 .

b. The amplitudes of the electric and magnetic fields at the surface, for a sinusoidal wave with intensity 330W/m2are500V/m and 1.7×10-6Trespectively.

Step by step solution

01

Define the intensity and define the formulas.

The power transported per unit area is known as the intensity .

The formula used to calculate the intensity ( l ) is:

l=PA

Where, A is area measured in the direction perpendicular to the energy and P is the power in watts.

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

Emax=2lε0CBmax=Emaxc

Where,ε0=8.85×10-12C/N.m2 and is the speed of light that is equal to 3.0×108m/s.

02

Determine the intensity of visible-light.

Given that,

P=75Wr=3×10-2m

Given that, only 5% of the energy goes to visible-light. So, we use the 5% of power ( P ) .

The formula used to calculate the intensity ( l ) is:

I=PA=5%P4π3×102=(0.05)×75113×104=331W/m2

Hence, the intensity of visible-light at the surface of the bulb330W/m2 .

03

Determine the amplitudes of electric and magnetic fields.

The amplitude of electric field is:

Emax=2Iε0c

Substitute the values

Emax=2×(331)8.85×10123×108=500V/m

The amplitude of magnetic field is:

Bmax=Emaxc

Substitute the values

Bmax=5003×108=1.7×106T

Hence, the amplitudes of the electric and magnetic fields at the surface, for a sinusoidal wave with intensity 330W/m2are 500 V/m and1.7×10-6T respectively.

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