Calculating Electric and Magnetic Field amplitude

AI Thread Summary
The discussion revolves around calculating the electric and magnetic field amplitudes of a circular laser beam with a power of 3.2 mW and a diameter of 8.5 mm. Participants successfully calculated the beam's intensity and the energy delivered over a short time but struggled with deriving the electric field amplitude from intensity. They discussed the relationship between electric and magnetic fields, emphasizing the need to use the Poynting vector and the time-averaged intensity to find the electric field amplitude. The conversation highlighted the importance of understanding wave properties and the mathematical relationships governing electromagnetic radiation. Ultimately, the group is working towards applying these principles to solve the problem effectively.
gsmtiger18
Messages
29
Reaction score
0

Homework Statement


A circular uniform laser beam has a power of P = 3.2 mW and a diameter of d = 8.5 mm. The beam is a form of electromagnetic raditation with both electric and magnetic field components.
(a) Calculate the intensity of the beam in units of W/m^2
(b) Calculate the energy (delta U) delivered in a time of t = .065 s in Joules.
(c) Use the intensity of the beam (I) to calculate the amplitude of the electric field in V/m
(d) The amplitudes of the electric and magnetic fields have a fixed relationship. Based on that relationship use the result of part (c) to calculate the amplitude of the magnetic field (B) in Tesla.I know that the power is P = 3.2 mW, and the diameter is d = 8.5 mm. From this I calculated intensity of the beam, so part (a) is complete. I can multiply that by time to get energy, so part (b) is complete as well. Where I'm running into a problem is somehow finding the electric field without knowing anything but the intensity and the energy in a small window of time. Typical calculations for electric potential energy haven't worked at all.

Homework Equations


delta U = Pt
I = P/A
A = pi*r^2
That's about all I know. Our professor left us totally on our own to learn this material.

The Attempt at a Solution


All I've managed is to calculate the delivered energy in a small window of time and the intensity of the beam.
 
Physics news on Phys.org
gsmtiger18 said:
(c) Use the intensity of the beam (I) to calculate the amplitude of the electric field in V/m
The intensity of a plane monochromatic wave is defined as the time average of Poynting vector. Do you the formula of Poynting vector?
 
S = 1/μ0E x B

However, how is this usable if I don't know both E and B?
 
gsmtiger18 said:
S = 1/μ0E x B

However, how is this usable if I don't know both E and B?
Good, now realize that ##S=|\mathbf{S}|## is a function of time. While, as I said in my previous post that the intensity is the time average of ##S## over one period of the wave,
$$
I = \langle S(t) \rangle_t
$$
So, first find out the how ##S## (the magnitude of ##\mathbf{S}##) looks like for a plane wave. (Note: you will also need to employ certain relation between the magnitude of the fields for a plane wave, with which you will solve part d)). Assume the electric field to have the form ##\mathbf{E} = \hat{x}E_0 \sin(\omega t - kz)##, employ Maxwell equations to find the corresponding ##\mathbf{B}##.
 
Last edited:
gsmtiger18 said:
Typical calculations for electric potential energy haven't worked at all.
There are no charges involved here, so there's no potential energy you can calculate.

gsmtiger18 said:
However, how is this usable if I don't know both E and B?
The magnitudes of the E and B fields are related.
 
vela said:
The magnitudes of the E and B fields are related.

That's what I keep finding in my text resources, but it's never elaborated, simply stated.

blue_leaf77 said:
So, first find out the how SSS (the magnitude of SS\mathbf{S}) looks like for a plane wave.

Would I be able to say that Em/c = Bm?
 
gsmtiger18 said:
Would I be able to say that Em/c = Bm?
That's correct, and I hope you know how to prove it, for which purpose the last sentence in part #4 was written.
So, now how does ##S## looks like in terms of ##E_m##?
 
I really have no idea how to prove the relationship I stated above, but in terms of Em, S = (1/μ0c) Em2
 
gsmtiger18 said:
I really have no idea how to prove the relationship I stated above
As the first step, follow what I wrote as the last sentence in post #4.
gsmtiger18 said:
S = (1/μ0c) Em2
Not really, you miss the time dependent behavior there. Look, the fields you have are
$$
\mathbf{E} = \hat{x}E_m \sin(\omega t -kz) \\
\mathbf{B} = \hat{y}\frac{E_m}{c} \sin(\omega t -kz)
$$
Plug in these equations into ##\mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}##.
 
  • #10
S = (1/μ0c)Em2sin2(ωt - kz)
 
  • #11
gsmtiger18 said:
S = (1/μ0c)Em2sin2(ωt - kz)
Yes, now take the time average of this function over one period ##T=2\pi/\omega##.
 
  • #12
What do I use for ω and z?
 
  • #13
gsmtiger18 said:
What do I use for ω and z?
##\omega## is used in relation to the period, while ##z## is actually irrelevant for the present calculation but it was introduced in order to make the expression for the field waves make sense, i.e. they represent propagating waves.
 
  • #14
I don't know the period.
 
  • #15
gsmtiger18 said:
I don't know the period.
Look it up in one of the previous posts of mine.
 
  • #16
You've lost me.
 
  • #17
I denoted the period as ##T##, in particular I also expressed it in terms of ##\omega##.
 
  • #18
So I can set ω = T/2π. Now what?
 
  • #19
gsmtiger18 said:
ω = T/2π
No, the right hand side should be the reciprocal of that one above.
gsmtiger18 said:
Now what?
What did I say in post #11?
 
  • #20
So I'd take the time average from 0 to T = 2π/ω?
 
  • #21
gsmtiger18 said:
So I'd take the time average from 0 to T = 2π/ω?
Yes.
 
  • #22
When I take that integral, would I include k as part of the sin2(ωt-kz) statement, since z is disregarded?
 
  • #23
gsmtiger18 said:
When I take that integral, would I include k as part of the sin2(ωt-kz) statement, since z is disregarded?
Do you mean the kz part in the argument of sine must be retained during the integral? Of course yes, it should be retained. Upon doing the integral, you should see that in the final result, the dependence on z (or k) is absence. Just compute the integral for now.
 
  • Like
Likes kamaljit
  • #24
This is a very interesting ques to me...we just had a class on this ☺
 
Back
Top