1. Dec 5, 2005

### ashkash

What is the magnitude of the force on the proton in the figure?

The figure is in the link below.

http://img207.imageshack.us/img207/5842/untitled9xb.png [Broken]

I have tried the Lorentz Force law that F = q(E + v x B) and got a numberical result of 4.8*10^-13 N, but that is wrong.

Last edited by a moderator: May 2, 2017
2. Dec 5, 2005

### dicerandom

What's the charge on the particle? And is that a 7 in the exponent for the velocity?

3. Dec 5, 2005

### ashkash

The charge on the particle is 1.6*10^-19, as it is a proton. Yes, that is a 7 in the exponent for the velocity.

the velocity is 2*10^7 m/s
E = 1.0 *10^6 V/m
B = 0.10 T

Sorry about the picture not being clear.

4. Dec 5, 2005

### dicerandom

Sorry, didn't see that you said it was a proton in your first post :)

If you are going to use the Lorentz force law here you'll have to use it in its full vector form:

$$\vec{F} = q\left(\vec{E} + \vec{v} \times \vec{B}\right)$$

If we call the direction to the right x, up y, and out of the page z, then you can see that the electric force will push the particle in the positive y direction whereas the magnetic force will push it in the positive x. The square of the magnitude of the force will then be the sum of the squares of the electric and magnetic forces (just from the Pythagorean Theorem).

5. Dec 5, 2005

### ashkash

thanks, it worked.

6. Nov 16, 2009

### astrydom

how did you find the force of the electric field on the proton to be up? I understand the reason for the direction of the magnetic force on the proton but not for the electric field. Would you be so kind as to explain please?