Does the Lorentz Force Affect Both the Wire and Loop in a Current Loop Setup?

[Samson4]

A current loop has a wire starting at its center. The wire terminates at the inside of the loop. The loop and wire have 150 and 15 ohms of resistance respectively. Both have 4 volts of potential across them.

Looking at this I understand the the wire will experience a lorentz force because the current inside it travels 90 degrees to the magnetic flux of the loop. But, what about the loop. Does the electrons in it not experience a lorentz force? Its not in the magnetic field of the wire. It must be equal and opposite right?

There is more current in the wire than the loop. Does this effect the forces each experience?

 

[BvU]

The electrons in the loop do indeed feel a Lorentz force from the magnetic field caused by the wire. From the geometry described the total force on the loop will be non-zero and in a direction along the length of the straight wire section. (wrt the straight wire section perpendicular components from left and right cancel. Longitudinal components from the nearer half-circle are a bit stronger than from the other half)

Field strengths -- and thereby forces -- are proportional to the currents, so: yes. And: the force, experienced by a current in a field is also proportional to the current. So yes: the product of both currents appears in each of the forces (loop on section, as well as section on loop).

 

[Samson4]

How would removing the wire and replacing it with a disc with radial currents from its center to its edge?

 

[BvU]

A conducting disk ? The current carriers would feel a kind of Coriolis force and be free to move in that direction.

 

[Samson4]

I just don't see what I'm missing in my understanding of the lorentz force and hall effect. For example; a pancake coil sitting atop a magnet. When current flows, the lorentz force should be into or out of the center of the coil. This is also the same direction of the current. So why doesn't current flow increase?

 

[BvU]

Because it can't go there. $\vec v\times \vec B$ points towards a boundary of the conductor. All that happens is the current carriers are pushed in or out a little bit. Charge buildup quickly compensates.

This is also the same direction of the current

No. $\vec v\times \vec B \perp \vec v$

 

[Samson4]

Because it can't go there. $\vec v\times \vec B$ points towards a boundary of the conductor. All that happens is the current carriers are pushed in or out a little bit. Charge buildup quickly compensates.

No. $\vec v\times \vec B \perp \vec v$

If you don't mind, I have one more question before I disappear into my studies. In regards to the current flow through a doped semiconductor. Does it behave more similar to the current flow though a vacuum than that of a conductor? Velocity doesn't seem to matter when discussing conductors. Calculations use amperage and forces are on the conductor as a whole.

 

[BvU]

Velocity doesn't seem to matter when discussing conductors. Calculations use amperage and forces are on the conductor as a whole

Think again. I is $dq\over dt$ and the amount of charge passing through a cross section with area A is $\rho v A$.

Semiconductors are like conductors, only the mobility of the charge carriers is a lot less.

 

[Samson4]

Think again. I is $dq\over dt$ and the amount of charge passing through a cross section with area A is $\rho v A$.

Semiconductors are like conductors, only the mobility of the charge carriers is a lot less.

If I drag a semiconductor through a magnetic field, it won't induce currents like a conductor would, right?

 

[BvU]

Why not ?

Semiconductors are like conductors, only the mobility of the charge carriers is a lot less.