Calculating Power Dissipation for a High Power Laser in Manufacturing

[akmphy]

Homework Statement

High power lasers in factories are used to cut through cloth and metal. One such laser has a beam diameter of 1.5mm and generates an electric field at the target having an amplitude of .854MV/m.
speed of light= 3.0e8 m/s
permeability of free space is 4pi x 10-7 TN/A
What is the power dissipated?

Homework Equations

P=SA
S= c(permeability of free space)E^2
Erms= 1/sqrt2(E)

The Attempt at a Solution

I first solved for the magnetic field, B= E/c, and got .002849T(this is correct).
I then solved for intensity, S, by calculating the R and B rms, and then using, S= c(permeability of free space)E^2. I got 9.6817e8 (this was correct).
I then calculated the power by using P= SA, and I got 6843.59 W.
What am I doing wrong? Thanks for any help

 

[gneill]

If your S is 9.68 x 10⁸ W/m² and your beam area 1.5mm square, then your power should be in the neighborhood of 2200W. Check your area conversion perhaps?

 

[akmphy]

So, S, intensity is 9.68e8, and the beam diameter is 0.0015m:
P = AS
= [(4pi)(.0015)^2][ S]
= 27369.55 W
I tried this one and it was wrong.
I tried dividing the diameter by two to get the radius:
P= AS
= [4pi (7.5e-4)^2]
= I got 6842.39
This was wrong.
I tried the area of a square, and not a circle shaped beam:
= .0015^2
= I got 2178W
Not correct.
Please help me with this. I already submitted the homework, but this is driving me crazy.
Thanks.

 

[gneill]

For a circular cross section, the area would be πr², or π(D/2)², no? Where is your factor of 4π coming from?

 

[akmphy]

For a circular cross section, the area would be πr², or π(D/2)², no? Where is your factor of 4π coming from?

I was approaching the laser like an imaginary sphere, so I think I was using the formula of 4pir^2. So, a cross sectional area...just an area of a circle?
P= 9.68e8 (.0015/2)2(pi)
I get 1710.59 W.
I already turned it in; is this the correct answer?
Thanks for your help

 

[gneill]

I was approaching the laser like an imaginary sphere, so I think I was using the formula of 4pir^2. So, a cross sectional area...just an area of a circle?
P= 9.68e8 (.0015/2)2(pi)
I get 1710.59 W.
I already turned it in; is this the correct answer?
Thanks for your help

Sorry, I don't know for sure. But it "feels right" to me.

 

[akmphy]

Sorry, I don't know for sure. But it "feels right" to me.

Thank you so much. I think it is right,too.