Forces between parallel currents

[georgia]

Homework Statement

Two narrow beams of electrons with velocities v and -v are injected into an evacuated chamber along the length l. Each beam moves with constant speed and carries a current I. The tendency for the beams to deflect each other through their mutual interaction is compensated by a magnetic flux density B perpendicular to the plane containing the two electron beams so that they travel in parallel straight lines a distance d apart (d << l). Show that the ratio f of the magnetic to the electric force on each of the beams due to their mutual interaction is vε0μ0

Homework Equations

For two parallel currents I1 and I2 the force on a length l of I2 is:

F= (μ0*I1*I2*l)/2πd

The Attempt at a Solution

Using the above equation and substituting I1 = I and I2 = -I:

F= -(μ0*I^2*l)/2πd

I don't know what to do now, what is the electric force?

All I know is Coloumb's law but I thought that only applied to static charges?

 

[Hootenanny]

Coloumb's law applies to all charges, stationary or not. However, in this case you need to consider the electric field of a finite line of charge. Have you met this before?

 

[m2003]

The electric force between two moving charges is given by the Lorentz force law, which states that the force F on a charged particle with charge q moving with velocity v in a magnetic field B is given by the equation:

F = q(v x B)

In this case, the beams of electrons are moving with opposite velocities v and -v, so the force on each beam will be in opposite directions:

F1 = -q(v x B)

F2 = q(-v x B)

Substituting in the value for the magnetic force from the given equation, we get:

F1 = -(μ0*I*v*B)/2πd

F2 = (μ0*I*v*B)/2πd

The net force on each beam will be the sum of the electric and magnetic forces:

F1 = -(μ0*I*v*B)/2πd + q(v x B)

F2 = (μ0*I*v*B)/2πd + q(-v x B)

Since the beams are moving with constant velocity, the electric force will be balanced by the magnetic force, so we can set the two equations equal to each other:

-(μ0*I*v*B)/2πd + q(v x B) = (μ0*I*v*B)/2πd + q(-v x B)

Simplifying and rearranging, we get:

2q(v x B) = (μ0*I*v*B)/πd

Since we know that the force on each beam due to their mutual interaction is equal to the force on the other beam, we can divide both sides by q:

2(v x B) = (μ0*I*v*B)/πd

Finally, we can substitute in the value for the electric field ε0, which is equal to 1/(μ0*c^2), where c is the speed of light:

2(v x B) = (v*ε0*μ0*I*v*B)/πd

Simplifying further, we get:

2(v x B) = (v*ε0*I*B)/πd

Dividing both sides by v, we get:

2(B/v) = (ε0*I*B)/πd

Finally, rearranging to solve for the ratio f of the magnetic to electric force, we get:

f = (B/v)/(ε0*I*B)/πd

f = v/(ε0*I)/πd

f = v