How Does a Capacitor Affect Conductor Motion in a Magnetic Field?

[zorro]

Homework Statement

This question is a product of my own mind.
The conductor of length l and mass m can slide without friction between two vertical conductors PQ and RS connected through a capacitor of capacitance C. A uniform magnetic field B is setup perpendicular to the plane (of figure). I am interested in finding out the motion of the conductor. We can assume ideal electric contact. The resistance of the wire QR is negligible.

By applying Newton's second law and the equation of capacitance Q=CV
I got the acceleration as a = g/(1 + B²l²C/m ).

As the induced current flows through the capacitor, it gets charged. A time comes when it blocks the current flow. But there will be an induced e.m.f. without any current flow. So there is no magnetic force on the conductor in upward direction.
During that time will the conductor fall through an acceleration g?
If yes then why is the acceleration obtained in my expression independent of current flow (constant)? Does the capacitor get discharged?

 

[yungman]

Homework Statement

This question is a product of my own mind.
The conductor of length l and mass m can slide without friction between two vertical conductors PQ and RS connected through a capacitor of capacitance C. A uniform magnetic field B is setup perpendicular to the plane (of figure). I am interested in finding out the motion of the conductor. We can assume ideal electric contact. The resistance of the wire QR is negligible.

By applying Newton's second law and the equation of capacitance Q=CV
I got the acceleration as a = g/(1 + B²l²C/m ).

As the induced current flows through the capacitor, it gets charged. A time comes when it blocks the current flow. But there will be an induced e.m.f. without any current flow. So there is no magnetic force on the conductor in upward direction.
During that time will the conductor fall through an acceleration g?
If yes then why is the acceleration obtained in my expression independent of current flow (constant)? Does the capacitor get discharged?

If it is according to your setup that you have a static mag field, there will be no current flow, there will be no Lorentz force in this case, nothing should move.

If you pulse with magnetic field, then Faraday's law apply where

$$V\;=\;-\int_C \vec E \cdot d\vec l = \int_S \frac {\partial \vec B }{ \partial t}$$

Where current will flow and you can play with Lorentz force.

Just a thought, I am no expert.

 

[zorro]

Well its not necessary that there should be a change in magnetic field.
There is a change in area enclosed by the loop formed thereby creating a change in magnetic flux. So induced current will flow.

 

[yungman]

Well its not necessary that there should be a change in magnetic field.
There is a change in area enclosed by the loop formed thereby creating a change in magnetic flux. So induced current will flow.

If there is no initial motion, why should there be change in total magnetic flux through the loop? If you give the wire a push, theroetically it won't stop anyway because there is no friction!

Also the way you drew, the magnetic field it in the middle of the loop only. If that is the case, increase the size of the loop will not increase the total flux.

Can you show me how to put a drawing like what you did? Thanks.

 

[zorro]

The conductor accelerates due to its weight acting downwards. Its understood.
Yes you are right about its motion, it doesnot stop - but there is a change in its acceleration.

The magnetic field exists uniformly throughout the region. I represented it using a circle and a cross. Its a way of representation if you don't want to draw crosses everywhere.

 

[zorro]

Some one help me out!

 

[aim1732]

Sorry for the late reply.
By your expression(which I have verified assuming segments PQ and RS to be of negligible resistance) the acceleration is constant and hence the current in the loop is constant and charge on the capacitor plates is increasing with time.

A time comes when it blocks the current flow

Why will that be? There is no theoretical limit on the charge a plate can bear---the value of E.dl will increase (which is q/c) but flux is increasing with time too. The only possible endgame for this is dielectric breakdown of air when your setup will collapse.
This is unrealistic because you have assumed that sections PQ and RS have no resistance. If they had it the current would have decreased with distance and that would have been a much more complex problem.

 

[zorro]

Why will that be? There is no theoretical limit on the charge a plate can bear---the value of E.dl will increase (which is q/c) but flux is increasing with time too. The only possible endgame for this is dielectric breakdown of air when your setup will collapse.

I read somewhere that a capacitor blocks DC current once it gets charged.

"the value of E.dl will increase (which is q/c) but flux is increasing with time too."

I did not get this :|

 

[aim1732]

Maxwell's Faraday's Equation: Integ[E.dl]=-d(Flux)/dt.
E is conservative and non conservative field resultant.

I read somewhere that a capacitor blocks DC current once it gets charged.

How is a capacitor charged?

 

[zorro]

By the initial flow of current - the source transfers the charges on the plates of the capacitor.

 

[aim1732]

Source here? How do you think it fixes the charge on capacitor?

 

[zorro]

Source here is the e.m.f produced due to induced electric field. We can replace the moving conductor by a battery.
The charges flow from this battery to the plates. This build up of charges on the plates creates an electric field which causes a potential drop across the capacitor. The charges/current continues to flow thill the potential across the capacitor equals the e.m.f. of the source.

 

[aim1732]

Yes yes you got it --- the emf increases with time and hence capacitor continues to charge infinitely.

 

[zorro]

Thanks alot.
That helped