Originally Posted by **fibonacci** given that the earth's magnetic field strength in perth is 5.8x10^-5 Wb/m^2
what is the max displacement of the beam spot due to earth expected if the accelerating volatge in an experiment is 300V? |

Hi there,

I think this may help. You can not find the max displacement of the beam spot without knowing how far the charge travels through the B field before striking the the return plate the beam of charge is directed at, but you can find the force that acts to displace the beam from its path of travel and then if you know the distance it travels you can find how far off the beam strikes the plate from where it was directed to hit the plate at.

You need to find the velocity of the charge to find the displacement force applied by the B field upon the charge. The voltage is related to the E field by V = Ed where d is the distance between the spot where the charge is first shot out of to the place where the beam strikes the return plate. So E = V/d = 300/d; and this is a contant valued E field. Now, F = Eq = ma, so Vq/d = ma, 300q/md = dv/dt.

(300q/md)dt = dv

(300qt/md) = V(t)

Now this is the instantaneous velocity as a function of time. You could go through and take the cross product of this value with the B field at every instant of time to find the force applied at those points in time (very difficult to do) or you can just assume that it moves at some average velocity, the total displacement found from either method should be the same; as while it is first moving slower than the average velocity it eventually exceeds the average velocity, such that the average velocity will yield the same displacement as using the instantaneous velocity at every point in time and finding all the different force vectors acting to push the charge a little more or less for every second of time that passes by.

The average velocity is the initial velocity summed with the final velocity divided by 2. The initial velocity is zero, so that means the average velocity is just the final velocity divided by 2. So:

V(t) = (300qt/md)

V(final) = V(t(final)) = (300qt(final)/md) = (300q/m)(t(final)/d)

= (300q/m)(1/Vave), now divide each side by 2 to get the average velocity from the final velocity, so:

Vave = V(final)/2 = (150q/m)(1/Vave) where (t(final)/d) = (1/Vave), because the total distance, d, divided by the total time it takes to travel that distance, t(final), is the average velocity.

Vave = (150q/m)(1/Vave)

Vave^2 = (150q/m)

Vave = sqrt(150q/m) where m is the mass of a charged particle within the beam.

The displacement force is now simply:

F = q(vxB) where v is the velocity, B is the magnetic field and q is the electric charge of the particle.

I'm pretty sure this is a valid method, my derivation of the average velocity is the only thing that concerns me, as I'm not positive it is correct, but it does make sense.

Now you have the average velocity, the B field is given, just look up the electric charge and mass for a particle shot off in the beam (most likely just an electron), and you'll have the force that will displace the beam by some amount depending on how long it travels through the B field, the longer it moves through the B field the more the beam is spatially displaced, the less time within the B field the less of a displacement of the beam.

Hope this helped.

Many Smiles,

Craig