Calculating Electric and Magnetic Field amplitude

[gsmtiger18]

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.

 

[blue_leaf77]

(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?

 

[gsmtiger18]

S = 1/μ₀E x B

However, how is this usable if I don't know both E and B?

 

[blue_leaf77]

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}##.

 

[vela]

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.

However, how is this usable if I don't know both E and B?

The magnitudes of the E and B fields are related.

 

[gsmtiger18]

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.

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 E_m/c = B_m?

 

[blue_leaf77]

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##?

 

[gsmtiger18]

I really have no idea how to prove the relationship I stated above, but in terms of E_m, S = (1/μ₀c) E_m²

 

[blue_leaf77]

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.

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}##.

 

[gsmtiger18]

S = (1/μ₀c)E_m²sin²(ωt - kz)

 

[blue_leaf77]

S = (1/μ₀c)E_m²sin²(ωt - kz)

Yes, now take the time average of this function over one period ##T=2\pi/\omega##.

 

[gsmtiger18]

What do I use for ω and z?

 

[blue_leaf77]

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.

 

[gsmtiger18]

I don't know the period.

 

[blue_leaf77]

I don't know the period.

Look it up in one of the previous posts of mine.

 

[gsmtiger18]

You've lost me.

 

[blue_leaf77]

I denoted the period as ##T##, in particular I also expressed it in terms of ##\omega##.

 

[gsmtiger18]

So I can set ω = T/2π. Now what?

 

[blue_leaf77]

ω = T/2π

No, the right hand side should be the reciprocal of that one above.

Now what?

What did I say in post #11?

 

[gsmtiger18]

So I'd take the time average from 0 to T = 2π/ω?

 

[blue_leaf77]

So I'd take the time average from 0 to T = 2π/ω?

Yes.

 

[gsmtiger18]

When I take that integral, would I include k as part of the sin²(ωt-kz) statement, since z is disregarded?

 

[blue_leaf77]

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.

 

[kamaljit]

This is a very interesting ques to me...we just had a class on this ☺