# Homework Help: Energy, magnitude of E & B, pressure from a laser

1. Apr 27, 2010

### darkfall13

1. The problem statement, all variables and given/known data

A Ti-sapphire femtosecond laser (1fs= 10^-15s, λ≈0.8μm) has an intensity of 1022 W/cm2
when focused to a spot of 1μm radius. What is the energy of the laser pulse? What is the peak
magnetic induction B in Tesla and the electric field in V/m? What is the wave pressure in
atmospheres?

2. Relevant equations

$$P = \int I da = \frac{E}{\Delta t}$$
$$E=cB$$

3. The attempt at a solution

I've solved the first part
$$P = IA = 10^{22} \frac{W}{cm^2} \pi \left(1 \mu m\right)^2$$
$$= 3.14 \times 10^{14} W$$
$$E = .314 J$$

But what is the relationship of what was given to either ]E or B? The only equations I can find are sinusoidal with the assumption that $$E_0$$ or $$B_0$$ are known to find their values at a particular time. Would I take the pulse with the assumption that B and E = 0 and the beginning and end and the information of the wavelength to find the value it peaks at? Now that I'm thinking of that, I'd still need some sort of E_0 or B_0.

Last edited: Apr 27, 2010
2. Apr 27, 2010

### BerryBoy

Assuming you have done the first part correct (if the question suggests it, you may need to integrate a gaussian distibution to get the power instead of just using the area of a circle)... anyway, to answer your question. The irrandiance $I$ relates to both the electric field $E$ and the magnetic induction $B$ through its definition as the time averaged Poynting vector:

$$I \equiv \left\langle S \right\rangle = \frac{c^2\epsilon_0}{2}\left\vert E_0 \times B_0 \right\vert$$

So that:
$$I = \frac{c}{2\mu_0}B_0^2$$
$$I = \frac{1}{2}c\epsilon_0 E_0^2$$

The "average" radiation pressure is given by the energy density of the wave (note the units energy/volume and force/area are the same)

$$P = \frac{1}{2}\epsilon_0 E_0^2$$ (this will give answer in Pascals though using previous units)

Hope this helps...

3. Apr 27, 2010

### darkfall13

That does help immensely thank you.