Question regarding Lorentz force.

[peripatein]

Hi,

Homework Statement

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.

Homework Equations

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) [itex]\varphi[/itex]^, whereas the correct answer, as asserted by the book, is (2πa²ρ₀/r)*(V/c)*(1 - e^-r2/a2) [itex]\varphi[/itex]^!
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.

 

[mfb]

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.

 

[sweet springs]

Hi.

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

 

[peripatein]

I realize 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?

 

[mfb]

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.

 

[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?

 

[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?

 

[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.

 

[mfb]

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

Isn't the net force equal to 0

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.