# Homework Help: Question regarding Lorentz force.

1. May 18, 2013

### peripatein

Hi,
1. The problem statement, all variables and given/known data
A system, with cylindrical symmetry and charge density ρ(r) = ρ0e-r2/a2, where a is a given constant, is given.
The system moves at a constant velocity V in the z^ direction. V<<c. The charge density in the lab's reference frame is approx. equal to that in the rest reference frame.
I am asked for the electrical and magnetic fields (magnitude and direction) everywhere.

2. Relevant equations

3. The attempt at a solution
I have managed to find the electrical field to be equal (2πρ0a2/r)*(1 - e-r2/a2) r^. I have confirmed that my answer is correct.
Now, what I don't quite understand is why couldn't I equate Lorentz force to zero ("constant velocity") in order to find the magnetic field? This would yield (2πa2ρ0/r)*(c/V)*(1 - e-r2/a2) $\varphi$^, whereas the correct answer, as asserted by the book, is (2πa2ρ0/r)*(V/c)*(1 - e-r2/a2) $\varphi$^!
What is the reason for this discrepancy?
By using Ampere's Law, I was able to solve it correctly, yet would like to understand why couldn't I simply have applied the requisite condition on Lorentz force to find the magnetic field?
I'd truly appreciate your help.

2. May 18, 2013

### Staff: Mentor

Lorentz force on what? In general, this system will lead to forces on all charges everywhere.
The net force on the system is zero.

I think I would perform a Lorentz transformation of the electric field of a non-moving system.

3. May 18, 2013

### sweet springs

Hi.

Your solution gives infinity when v=0. It is not reasonable at all.

4. May 18, 2013

### peripatein

I realise that my solution is wrong. My question was simply: why could I not have used F = q(E + (V/c)xB) = 0, since it is stated that the system moves at a constant speed<<c and in section A of that same question I found the magnitude of the electrical field. Why couldn't I simply substitute that into the above equation for the net force therefrom to find B?

5. May 18, 2013

### Staff: Mentor

The system does not consist of isolated charges, magically held in their position by the Lorentz force (that does not work). To realize such a charge distribution, you would need a perfect insulator, where charges keep their position independent of electric and magnetic fields.

6. May 18, 2013

### sweet springs

Hi.

Lorentz force comes from electric filed E and magnetic field B. Even when velocity of the cylinder is zero, there exists electric force. B generated by the slow speed of cylinder is proportional to that speed, not v, in the first order approximation. v is the speed of charge we put on the position we are dealing with so we can set it in any value we like, not a fixed or constant value. So how the idea F=0 came to you?

Last edited: May 18, 2013
7. May 18, 2013

### peripatein

I am not sure I understand. Isn't the velocity in the formula for the force that of the cylinder? And if so, why couldn't that formula be applied to find B once I know E? Isn't the net force equal to 0, since the cylinder is moving with a constant velocity?

8. May 18, 2013

### sweet springs

Hi.

V is the speed of the cylinder.
v in your Lorentz formula is the speed of the test charge that is variable, or you can say electric current there.
For example if you put the test charge still, i.e. v=0, the charge receives electric field E but no magnetic field force because vXB=0 even for non zero B.

Last edited: May 18, 2013
9. May 18, 2013

### Staff: Mentor

Just ignore forces, they are not relevant in this problem.

The net forces on charges in the setup is not zero. The net force along the cylindrical symmetries is 0, but you still get a net force in radial direction.