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

#### neilparker62

Homework Helper
Deleting my post. Have realised 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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#### hutchphd

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

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

#### rude man

Homework Helper
Gold Member
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 = $2\pi(\frac {m + b^2d\epsilon_0 B^2} k)^{1/2}$

#### Michael Price

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 = $2\pi(\frac {m + b^2d\epsilon_0 B^2} k)^{1/2}$
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.

#### hutchphd

2 lines and 25 mutually contradictory posts.

#### neilparker62

Homework Helper
Cool Physics all round - thanks all!

#### Michael Price

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

#### Moara

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

#### hutchphd

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

#### TSny

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

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.
I was never thinking in terms of going to the frame of the plate. I was staying in the lab frame where the important forms of energy of the system are kinetic energy, potential energy of the spring, and the field energy of the electric field E between the square surfaces of the plate.

In fact it will change the internal energy of copper by separating charges onto the surfaces.
The energy of the electric field between the surfaces of the plate accounts for the energy associated with separating the charge.

The energy stored by those charges is then

Uint0E2b2d/2

0(v2/c2)B2b2d/2​
I agree with these expressions. But it is important to keep in mind that this is the energy associated with the E field, not the B field. It just happens that this electric field energy can be expressed in terms of $B$ and $v$.
∴ meff=UB/c2
I don’t see how you get this.The extra effective mass has the value $m_{\rm eff} = \varepsilon_0 B^2 b^2 d$ corresponding to answer (c). This may be written as $2U_B/c^2$. It is interesting that this “extra mass” of the system can be written as twice the mass equivalence of the magnetic field energy inside the plate. But, I think it is misleading to think of the extra mass term as actually due to the mass equivalence of the B field energy. The important energy is the electric potential energy of the charge that accumulates on the surfaces of the plate, which is given by the E field energy. This energy is proportional to $v^2$ and thus effectively acts like an extra kinetic energy with mass $m_{\rm eff}$.

[Edited]

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#### hutchphd

I don't think we disagree about the physics. I simply point out that

UB=(volume) B2/u0
0c2(b2d) B2
There is no factor of 2 required (?) .

Also I like to think the energy in question is the energy required to exclude the E field from the conductor via the charge separation.

#### TSny

Homework Helper
Gold Member
I don't think we disagree about the physics. I simply point out that

UB=(volume) B2/u0
0c2(b2d) B2
There is no factor of 2 required (?) .
There’s a factor of 2 because the energy density of the magnetic field has a factor of 1/2:
$u_B = \frac{1}{2}\frac{B^2}{\mu_0}$.

There’s a similar factor of 1/2 in the energy density of the E field.
Also I like to think the energy in question is the energy required to exclude the E field from the conductor via the charge separation.
In the lab frame the E field will not be excluded from the interior of the plate. The E field inside the copper creates an electric force on the free electrons that essentially cancels the magnetic force.

But I think you’re considering things in a reference frame moving with the plate. When you switch to that frame, the net E field inside the plate will be zero. The E field produced by the charge separation cancels the E field in this frame that comes from the transformation laws for the fields when switching frames of reference. That is, the B field in the lab frame is seen as both a B field and an E field in the the plate frame. The charge separation then creates an E field that makes the total E equal to zero in the plate frame.

#### hutchphd

There’s a factor of 2 because the energy density of the magnetic field has a factor of 1/2:
$u_B = \frac{1}{2}\frac{B^2}{\mu_0}$.

There’s a similar factor of 1/2 in the energy density of the E field.
In the lab frame the E field will not be excluded from the interior of the plate. The E field inside the copper creates an electric force on the free electrons that essentially cancels the magnetic force.

But I think you’re considering things in a reference frame moving with the plate. When you switch to that frame, the net E field inside the plate will be zero. The E field produced by the charge separation cancels the E field in this frame that comes from the transformation laws for the fields when switching frames of reference. That is, the B field in the lab frame is seen as both a B field and an E field in the the plate frame. The charge separation then creates an E field that makes the total E equal to zero in the plate frame.
1/2

How can I say this succinctly.......oops, my bad. Thanks for the correction and apologies for wasting your time!!!

#### TSny

Homework Helper
Gold Member
1/2

How can I say this succinctly.......oops, my bad. Thanks for the correction and apologies for wasting your time!!!
No apologies needed. It’s an interesting problem.

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

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