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

[TSny]

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

U_int =ε₀E²b²d/2

=ε₀(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 $v$.

∴ m_eff=U_B/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]

 

[hutchphd]

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

U_B=(volume) B²/u₀
=ε₀c²(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]

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

U_B=(volume) B²/u₀
=ε₀c²(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.

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

 

[TSny]

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.