Eletromagnetism: Copper Plate on a Spring Oscillating in a Magnetic Field

• Moara
In summary, the conversation discusses the use of Faraday's law to find the difference of potentials in a plate and the effect of a magnetic field on the plate. It is determined that there will be a "motional emf" in the plate equal to B.d.v, where v is the vertical velocity of the plate. The concept of surface charge distribution and the resulting current and magnetic force is also mentioned. The effect of copper conductivity on the motion of the plate is discussed, with the conclusion that for good conductors, the answer is independent of the conductivity. Finally, the concept of effective mass is brought up in relation to the motion of the plate.
Moara
Homework Statement
A plate of cupper with mass m has a square shape with side length b and thickness d. The plate is suspended vertically by a spring with constant k in an uniform magnetic field horizontal and parallel with the plates plane. Find the period of vertical oscillations of the plate
Relevant Equations
Fel=kx, U=d(flux)/dt, T=2pi √m/keq
Tried to find the resultant force, but I can't see how the magnetic field affects. I used Faraday's law to find the the diferece of potentials in the plate Wich should be B.d.v, where v is the vertical velocity of plate, but there were not given the resistance or resistivity to relate with the current and afterwards with the magnetic force

Last edited by a moderator:
Moara said:
I used Faraday's law to find the the diferece of potentials in the plate Wich should be B.d.v, where v is the vertical velocity of plate, but there were not given the resistance or resistivity to relate with the current and afterwards with the magnetic force
Yes, there will be a "motional emf" in the plate equal to ##B \, d \, v##. Can you find the electric field ##E## associated with this emf? Can you find the surface charge distribution ##\sigma## on the plate that produces the electric field?

The general idea is that as the plate changes its speed, ##\sigma## will change. There will be a current associated with the change in ##\sigma##. The magnetic field causes a force on the current.

Another approach is to use energy concepts. There is an energy associated with the electric field induced in the plate. If you include this energy as part of the total energy of the system, then you can deduce the answer from the form of the total energy.

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Thank you

rude man said:
Seems to me the conductivity σ of copper should appear in T. After all, if σ is low there is no effect by B, and if σ → ∞ there would be no motion at all.
I don't think that ##\sigma_c \rightarrow \infty## would mean that the plate would not be able to move. (I put a subscript "c" here to distinguish the conductivity from the surface charge density ##\sigma## on the plate.) An infinite ##\sigma_c## would mean that the free charge in the copper would be able to instantaneously redistribute itself when perturbed. But this just guarantees that the charge density on the plate will always be just what is necessary to produce the value of the electric field ##E = Bv## associated with the motional emf as ##v## changes during the oscillation of the plate.

The time it takes for free charge in a conductor to redistribute itself is on the order of the "relaxation time", which is inversely proportional to ##\sigma_c##. Copper's large value of ##\sigma_c## yields a relaxation time that is extremely small, something like 10-14 s. This time is so much shorter than any reasonable period of oscillation of the plate that you can take ##\sigma_c## for copper to be essentially infinite for this problem.

I know very little about superconductivity. But I don't believe that infinite ##\sigma_c## necessarily implies superconductivity, although the converse is true.

See https://en.wikipedia.org/wiki/Perfect_conductor

I will be traveling for about a week starting tomorrow. I probably won't be able to respond until I get back. Maybe others will join the discussion in the meantime.

rude man said:
If conductivity were infinite then the tiniest dB/dt would generate infinite eddy currents and infinite restoring force. Therefore, no vertical motion. This is how maglev trains work, using superconducting rails, as I understand it.
The eddy currents generated in a superconductor are sufficient to completely exclude the external field. Not infinite

Do this the easy way!:

What if B=0? (eliminates a,b)
Will B make T longer or shorter? (eliminates d)
If m=0 it will not oscillate (eliminates a,b,c)
Sorry, what is answer e? ...

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hutchphd said:
Do this the easy way!:

What if B=0? (eliminates a,b)
Will B make T longer or shorter? (eliminates d)
If m=0 it will not oscillate (eliminates a,b,c)
Sorry, what is answer e? ...
The term ##\varepsilon_0 B^2 b^2 d## acts like an effective mass. So the plate could still oscillate even if the true mass ##m## is zero. But I agree that you have nicely eliminated some of the answers.

rude man said:
"In its superconducting state the wire has no electrical resistance and therefore can conduct much larger electric currents than ordinary wire, creating intense magnetic fields. "

https://en.wikipedia.org/wiki/Superconducting_magnet
I don’t think we need to worry about the infinite conductivity case. If the plate were a poor conductor so that the free electrons are “sluggish”, then that would modify the analysis in a complicated way that would depend on the value of the conductivity. But these complications are negligible for good conductors. So, the answer is independent of the conductivity to a very good approximation.
Have a good trip T.
Thanks! I’m currently killing time at an airport.

TSny said:
The term ε0B2b2dε0B2b2d\varepsilon_0 B^2 b^2 d acts like an effective mass
An effective mass would multiply the acceleration not the velocity.

hutchphd said:
An effective mass would multiply the acceleration not the velocity.
The effect of the B field is to produce a force on the plate that is proportional to the acceleration of the plate. The coefficient of proportionality is the effective mass and is expressed by one of the B field terms in the answers.

How big is this effect?
My quick analysis says it is egual to the mass equivalent of the field energy of the B field inside the copper! Has this actually been done?
I am dubious but open to enlightenment. Please.

Michael Price
Moara said:
Problem Statement: A plate of cupper with mass m has a square shape with side length b and thickness d. The plate is suspended vertically by a spring with constant k in an uniform magnetic field horizontal and parallel with the plates plane. Find the period of vertical oscillations of the plate
Relevant Equations: Fel=kx, U=d(flux)/dt, T=2pi √m/keq

Tried to find the resultant force, but I can't see how the magnetic field affects. I used Faraday's law to find the the diferece of potentials in the plate Wich should be B.d.v, where v is the vertical velocity of plate, but there were not given the resistance or resistivity to relate with the current and afterwards with the magnetic forceView attachment 246068
Since the answers are all different even if the B field is zero, set the B to zero and solve.
Answer is c. Or e.

hutchphd said:
How big is this effect?
It is very small for any practical value of B. The ratio of the effective mass term to the mass of the plate is ##\varepsilon_0 B^2/\rho##, where ##\rho## is the mass density of copper. So the effect on the period of motion is too small to measure. The problem seems to be an interesting academic exercise.
My quick analysis says it is egual to the mass equivalent of the field energy of the B field inside the copper!
Yes, nice observation!

## \varepsilon_0 B^2b^2d = 2 U/c^2##, where U is the B-field energy in the plate and ##c## is the speed of light in vacuum.

Michael Price
Michael Price said:
Since the answers are all different even if the B field is zero, set the B to zero and solve.
Answer is c. Or e.
Yes

Thanks for the confirmation. If given as an elementary problem, I find it to be misguided, particularly when it ignores the really useful viscous drag which, even in this geometry, will dominate in the real world. I hope @Moara understands this.

hutchphd said:
How big is this effect?
My quick analysis says it is egual to the mass equivalent of the field energy of the B field inside the copper! Has this actually been done?
I am dubious but open to enlightenment. Please.
Yes! I get the mass equivalent of twice the B field energy, presumably because there is an equal amount of energy stored in the E field. A really neat result. So 'c' is the answer.

Michael Price said:
Yes! I get the mass equivalent of twice the B field energy, presumably because there is an equal amount of energy stored in the E field. A really neat result. So 'c' is the answer.
The relation between the E-field energy ##U_E## and the B-field energy ##U_B## is

##U_E = \frac{v^2}{c^2} U_B##

where ##v## is the instantaneous speed of the plate. This may be rewritten as

##U_E = \frac{1}{2} m_{eff} v^2## where ##m_{eff} = 2U_B/c^2##.

So the electric field energy acts like an extra contribution to the kinetic energy.

Michael Price
T,
after further thought I'm thinking (a) there are no currents anywhere so my previous posts were irrelevant; (b) there
are Em and Es fields, equal & opposite, = Bvd, , all over the plate's insides, pointing front to back & reverse depending on mass motion direction. Tbis implies static charges all over the back and front of the plate also alternating with said motion.

I've read your posted energy argument but I can't see physically how that works. Something must apply force to the mass somehow, other than kx or mg. Don't see the charges in this. v x B is in the wrong direction

I'm hoping "N.R.A." is Portuguese for "none of the above".

As the charges migrate (to the big flat faces) vmigrate x B will be along the y direction?
I do not like this problem, particularly for pedagogy. For instance I bet the temperature change caused by the real world finite conductivity causes a far larger mass change.

hutchphd said:
As the charges migrate (to the big flat faces) vmigrate x B will be along the y direction?
That may be the answer! Certainly something to think about. Thx.

Excuse me, at first I thought I could solve this easily but after a careful look I discovered there are some things that I really can't understand. I'd like your help.

1) (this might be stupid) If ##B## is uniform everywhere, how is it possible to have an emf ? The change in flux through the lateral surface would be zero.

2) Let's say there is this some emf, then how do I quantify the current flowing from one face to the other if I don't know the conductivity ? If I don't know the current how can I quantify the force and thus find the period ?

3) Can someone explain better how to link the energy of the field to the period of oscillations ?

Sorry for the interruption. I'd be very happy if someone could answer me.

Thanks
Ric

dRic2 said:
Excuse me, at first I thought I could solve this easily but after a careful look I discovered there are some things that I really can't understand. I'd like your help.

1) (this might be stupid) If ##B## is uniform everywhere, how is it possible to have an emf ? The change in flux through the lateral surface would be zero.

2) Let's say there is this some emf, then how do I quantify the current flowing from one face to the other if I don't know the conductivity ? If I don't know the current how can I quantify the force and thus find the period ?

3) Can someone explain better how to link the energy of the field to the period of oscillations ?

Sorry for the interruption. I'd be very happy if someone could answer me.

Thanks
Ric
@dRic2,
i'll give you my shot. Don't be alarmed; I think everyone here so far has some uncertainty about it.
I think post 23 may have the best answer. As I see it there is one (actually two, see below) electric field between the large sides. This is from emf = Ed = Bvd or E = Bv. The vector version tells you the E field is directed between the large surfaces thru d.

I don't know if you understand the distinction between electrostatic E fields (Es) and induced (emf) ones (Em). Here we have one Es and an equal and opposite Em (there can be no net E field in the copper). Em = Bv and so Es = -Bv. The Es is the field associated with (beginning and ending in) charges. These charges exist on both large exterior surfaces and alternate +/- then -/+ as the mass moves up & down with v. This motion forms an alternating current thru the d layer which then is acted upon by the B field by the F = BIL rule. The vector version dictates that the force is up & down.

At the moment this is the best I've come across. We're all hoping others will join in, giving their respective interpretations. Anyway, it's certainly not a trivial problem IMO.

dRic2
dRic2 said:
1) (this might be stupid) If BBB is uniform everywhere, how is it possible to have an emf ? The change in flux through the lateral surface would be zero.

@TSny post #21 gives the E field seen by a moving observer (on the plate) relative to a B field (in the "lab"). Alternatively you could talk about the Hall Effect charge separation on the flat plate caused by its motion. Same physics, different viewpoint. In any event charge is separated and that energy can be calculated (surface charge on parallel plates). It turns out to be proportional to B2 and the volume and (v2/c2)3) Can someone explain better how to link the energy of the field to the period of oscillations ?
Let U be this energy (stored now in charge separated). dU/dy = will be the associated force as y changes and since U depends on vy we write this as
dU/dy = (ay /vy )(dU/dvy )
=meffay
And mass renormalization takes care of it. The effect is very very small.

Yes, nice observation!
## \varepsilon_0 B^2b^2d = 2 U/c^2##, where U is the B-field energy in the plate and ##c## is the speed of light in vacuum.
[/QUOTE]
So now we're doing relativity! Judging by the very tiny change in mass (I got ~1e-17kg for a 0.01T B field, d=1cm, b=10cm.) so even for a light mass of 1gm that's ~1e-14 change. Have to admit, had I done that calculation at the outset I would never have proceeded. Thanks to you guys coming up with at least a plausible derivation. Far-out problem!

It's in Portuguese so good chance it's Brazilian. Some years ago Feynman was asked to look at Brazilian physics textbooks & lambasted the locals for a total absence of quantitatively meaningful problems. Maybe the
locals took his advice to heart!

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Michael Price
Deleting my post. Have realized this problem is far more complicated than it appeared at first glance. Better leave it to T.Sny et al !

Re query on meaning of option E:

Asking "Uncle Google" to translate "none of the answers above" I get "nenhuma das respostas acima" and I guess NRA would be the abbreviation for that.

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Thanks for translation.
There is a factor of 2 floating around in this answer vis

TSny said:
The relation between the E-field energy ##U_E## and the B-field energy ##U_B## is

##U_E = \frac{v^2}{c^2} U_B##

where ##v## is the instantaneous speed of the plate. This may be rewritten as

##U_E = \frac{1}{2} m_{eff} v^2## where ##m_{eff} = 2U_B/c^2##.

So the electric field energy acts like an extra contribution to the kinetic energy.
I do not think this is correct.
In the frame of the copper, the E field is external and cannot simply be added to the KE by fiat. In fact it will change the internal energy of copper by separating charges onto the surfaces. The energy stored by those charges is then

Uint0E2b2d/2

0(v2/c2)B2b2d/2

∴ meff=UB/c2

I see no need for relativistic handwaving.
If we follow the lines of reasoning in posts 23 and 26:
## E_s = Bv ##
D = ## \epsilon_0 E_s = \epsilon_0 Bv ##
## \sigma = D ## where σ = surface charge density
Q = σA = ## \epsilon_0 Bvb^2 ##
i = dQ/dt = ## \epsilon_0 Bb^2 dv/dt ##
Now the equation of motion becomes
## m\ddot y = -mg -ky - Bid ##
but ## dv/dt = d^2y/dt^2 ## so the equation of motion becomes
## (m + B^2d\epsilon_0b^2)\ddot y + ky = -mg ##
and the obvious effect of B is to change ##\omega^2 ## from the usual k/m to
## \omega^2 = \frac k {(m + B^2d\epsilon_0b^2)} ##
with the new period T = 2π/ω = ## 2\pi(\frac {m + b^2d\epsilon_0 B^2} k)^{1/2} ##
so the answer is (c).

neilparker62 and dRic2
rude man said:
I see no need for relativistic handwaving.
If we follow the lines of reasoning in posts 23 and 26:
## E_s = Bv ##
D = ## \epsilon_0 E_s = \epsilon_0 Bv ##
## \sigma = D ## where σ = surface charge density
Q = σA = ## \epsilon_0 Bvb^2 ##
i = dQ/dt = ## \epsilon_0 Bb^2 dv/dt ##
Now the equation of motion becomes
## m\ddot y = -mg -ky - Bid ##
but ## dv/dt = d^2y/dt^2 ## so the equation of motion becomes
## (m + B^2d\epsilon_0b^2)\ddot y + ky = -mg ##
and the obvious effect of B is to change ##\omega^2 ## from the usual k/m to
## \omega^2 = \frac k {(m + B^2d\epsilon_0b^2)} ##
with the new period T = 2π/ω = ## 2\pi(\frac {m + b^2d\epsilon_0 B^2} k)^{1/2} ##
so the answer is (c).
I did it slightly differently. I worked out the drag for arbitrary acceleration (not just SHM), and then F/a is the increased effective mass. No need for relativity, again - except that Maxwell's equations are relativity.

hutchphd
But by understanding relativity one can realize
1. The effect will be of order (v2 /c2 )
2. The correct solution takes two lines using total energy conservation.
Your choice...

rude man said:
2 lines and 25 mutually contradictory posts.
I never contradicted myself once.
Your choice...

Cool Physics all round - thanks all!

rude man said:
Maxwell's equations hold in relativity but obviously were formulated before relativity was introduced. This is strictly a classical physics problem.
Classical physics is often taken to include special relativity simply because Maxwell's equations imply SR. It just took Lorentz, Poincare, Fitzgerald,... - and eventually Einstein - 40 years to work this out.

Electrostatics + SR => magnetism
Or
Electromagnetism => SR

Take your pick.

In my book the answer is c. I think that, although physically incomplete or purely theoretical, the problem wants to handle only with some "easier" observations of the energy associated with B and E and the induced charges. By the way, I don't know where it is from, but does not come from Brazil. It might be from some olympiad or Russian book

Agreed...it was fun. Hope you learned something.

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