A falling resistive disk in uniform magnetic field

[AmanWithoutAscarf]

Homework Statement

I'm trying to figure out the exact current between the two plates of a disk with resistivity p, mass m, radius R and thickness d, falling in a uniform magnetic field that are parallel to the disk's surface and perdipencular to gravity field.

Relevant Equations

$\displaystyle J=\frac{1}{\rho } .\mathbf{E}$

The original question was to find the final velocity after falling from the height $H$

In many correct solutions to this problem, they consider the current density between the two surfaces to be $\displaystyle J=\frac{1}{\rho } .\mathbf{E^{*}} =\frac{\mathbf{v} \times \mathbf{B}}{\rho }$

But I think there must be an internal electric field $E_i$ inside the disk, caused by the surface charge density, because the charges cannot change immediately due to resistance. So, electric current would be $\displaystyle J=\frac{1}{\rho } .(\mathbf{v \times B-E}_{i})$

Is there anything that was mistaken?

 

[haruspex]

I'm not quite clear what question you are trying to solve.
As you note, there will be a build up of charge, slowing the current.
Are you looking for the general equation of how that builds up as the disc accelerates?
What about any reduction in that acceleration that results?

 

[AmanWithoutAscarf]

Ah sorry for the confusion, I'm just trying to figure out the electric current inside the disk of that electrodynamic problem, but it seems that I got it wrong compared to the correct solution and I don't know why.

 

[TSny]

@AmanWithoutAscarf
Can you give us the complete statement of the problem as originally given to you?

 

[haruspex]

"perdipencular"

That does deserve to be a word.

 

[AmanWithoutAscarf]

@AmanWithoutAscarf
Can you give us the complete statement of the problem as originally given to you?

Yes.
There is a conducting disk with resistivity p, mass m, radius R, and thickness d, falling in a uniform magnetic field from a height of $H$ (the field is parallel to the disk's surface and perpendicular to the gravity field, as illustrated in the image). Find the final velocity of the disk when it reaches the ground.
Sorry if there is any language mistake.

 

[TSny]

Yes.
There is a conducting disk with resistivity p, mass m, radius R, and thickness d, falling in a uniform magnetic field from a height of $H$ (the field is parallel to the disk's surface and perpendicular to the gravity field, as illustrated in the image). Find the final velocity of the disk when it reaches the ground.
Sorry if there is any language mistake.

OK. Thank you.

In many correct solutions to this problem, they consider the current density between the two surfaces to be $\displaystyle J=\frac{1}{\rho } .\mathbf{E^{*}} =\frac{\mathbf{v} \times \mathbf{B}}{\rho }$

But I think there must be an internal electric field $E_i$ inside the disk, caused by the surface charge density, because the charges cannot change immediately due to resistance. So, electric current would be $\displaystyle J=\frac{1}{\rho } .(\mathbf{v \times B-E}_{i})$

I agree with you. The internal electric field $E_i$ needs to be taken into account.

$E_i = \sigma/\varepsilon_0$ and $J = \dot \sigma$.

I believe you can set up two coupled, first-order differential equations for $\sigma(t)$ and $v(t)$, where $v(t)$ is the speed of fall of the disk. When I do this and solve for $v(t)$, I get that $v(t) \approx g t$ for any reasonable choice of values for $B$, $\rho$, $R$, $d$, and $m$. So, the disk will essentially fall with constant acceleration $g$.

 

[kuruman]

So, the disk will essentially fall with constant acceleration $g$.

I have a different point of view because I cannot imagine a conductor falling in a magnetic field without reaching terminal velocity because of magnetic braking. For simplicity, I considered a rod of length $L$ equal to the thickness of the disc and oriented parallel to the axis of the disc. The disc in this case can be assembled as a collection of rods.

I define Cartesian axes with $\mathbf{\hat x}$ along the field, $\mathbf{\hat y}$ along the rod and $\mathbf{\hat z}$ along the acceleration of gravity (down). We know that if a rod moves with constant velocity $\mathbf{v}$ in a direction perpendicular to a uniform magnetic field $\mathbf{B}$, there will be charge separation at the two ends and an emf across the rod given by $\text{emf}=BLv$.

However, if there is acceleration, the amount of charge at each end will increase and keep increasing which means current flow. If the resistance of the rod is $R$, the induced current in the rod will be $$I_{\text{ind}}=\frac{BLv}{R}$$ and there will be a Lorentz force opposing the velocity $$\mathbf{F}_M=I \mathbf L \times \mathbf B=-\frac{B^2L^2v}{R}\mathbf {\hat z}.$$Then the EOM to solve is $$m\frac{dv}{dt}=mg-\frac{B^2L^2v}{R}.$$The terminal velocity is $$v_{\text{ter.}}=\frac{mgR}{B^2L^2}.$$ When the rod reaches this velocity, we have steady state and zero acceleration.

 

[TSny]

I have a different point of view because I cannot imagine a conductor falling in a magnetic field without reaching terminal velocity because of magnetic braking. For simplicity, I considered a rod of length $L$ equal to the thickness of the disc and oriented parallel to the axis of the disc. The disc in this case can be assembled as a collection of rods.

I define Cartesian axes with $\mathbf{\hat x}$ along the field, $\mathbf{\hat y}$ along the rod and $\mathbf{\hat z}$ along the acceleration of gravity (down). We know that if a rod moves with constant velocity $\mathbf{v}$ in a direction perpendicular to a uniform magnetic field $\mathbf{B}$, there will be charge separation at the two ends and an emf across the rod given by $\text{emf}=BLv$.

Yes. There will not be any current in this case since the magnetic force on a charge carrier is canceled by the electric force from the charge accumulation at the ends of the rod.

However, if there is acceleration, the amount of charge at each end will increase and keep increasing which means current flow.

However the current will be very small since the magnetic and electric forces on a charge carrier in the rod will almost cancel while the rod is accelerating.

If the resistance of the rod is $R$, the induced current in the rod will be $$I_{\text{ind}}=\frac{BLv}{R}$$

I don't think this expression for the current takes into account the near cancellation of the electric and magnetic forces. I might be mistaken, but I think this expression would correspond to the rod (or disk) making up part of a complete circuit.

For example, we could imagine the rod (disk) sliding vertically between two conducting plates such that the plates are connected at the bottom. The disk makes contact with the plates with negligible friction. So, we have a complete circuit for current to flow around. Now we would get a terminal velocity as you calculated.

 

[haruspex]

With positive charge built up on one face and negative charge on the other, won't there be a torque causing the disc to rotate to be parallel to the field?

 

[kuruman]

I don't think this expression for the current takes into account the near cancellation of the electric and magnetic forces.

Now that I'm rethinking all this I'm probably wrong because I'm double counting the Lorentz force. However, it bugs me that the force balance $qvB=qE\implies E=vB$ says that the rod in free fall has increasing velocity which means increasing magnetic force which means increasing electric force which means increasing charge redistribution.

 

[TSny]

With positive charge built up on one face and negative charge on the other, won't there be a torque causing the disc to rotate to be parallel to the field?

My right hand is telling me that the forces on the faces are in the direction to cause the faces to repel each other without any torque on the disk.

 

[haruspex]

My right hand is telling me that the forces on the faces are in the direction to cause the faces to repel each other without any torque on the disk.

Sorry, I didn’t look at the diagram carefully enough. I thought the field was parallel to the axis of the disk.

 

[TSny]

However, it bugs me that the force balance $qvB=qE\implies E=vB$ says that the rod in free fall has increasing velocity which means increasing magnetic force which means increasing electric force which means increasing charge redistribution.

I think that's right. Of course, for the charge accumulation to increase as the rod accelerates, there must be a small current. So, at any time, qvB must be a little bit greater than E.

 

[kuruman]

My right hand is telling me that the forces on the faces are in the direction to cause the faces to repel each other without any torque on the disk.

An equivalent way to see it is in the moving frame of the disc.

When the disc is moving with constant velocity $\mathbf v$ relative to the lab frame in which the magnetic field is at rest, the relativistic field transformation equations say that in the rest frame of the disc there is only a uniform electric field perpendicular to the velocity given by $$\mathbf E'_{\perp}=\gamma \mathbf v \times \mathbf B$$. At non-relativistic speeds, as is the case here, $\gamma \approx 1$. Thus, an observer at rest with the disc will see an external uniform field. In this field there will be induced surface charges at each face that will cancel the electric field inside the disc. Thus, the disc constitutes a dipole in a uniform external electric field. The dipole moment is in the same direction as the external electric field that induced it so there will be no torque $\vec \tau=\vec p \times \vec E$ and the potential energy $U=-\vec p \cdot \vec E$ is at a minimum.

I am not sure how the picture changes when the velocity increases as a function of time.

 

[AmanWithoutAscarf]

OK. Thank you.


I agree with you. The internal electric field $E_i$ needs to be taken into account.

$E_i = \sigma/\varepsilon_0$ and $J = \dot \sigma$.

I believe you can set up two coupled, first-order differential equations for $\sigma(t)$ and $v(t)$, where $v(t)$ is the speed of fall of the disk. When I do this and solve for $v(t)$, I get that $v(t) \approx g t$ for any reasonable choice of values for $B$, $\rho$, $R$, $d$, and $m$. So, the disk will essentially fall with constant acceleration $g$.

I did try to solve the differential equations.
From the given assumptions and the second Newton law, we can set up two equations:
$$\dot{q} =\frac{S}{\rho }\left( v.B-\frac{q}{S.e}\right)$$ (1)
$$\dot{v} =g-\frac{B.d.\dot{q}}{m}$$ (2)
Take the derivative of (1) over time, then substitute to (2), we have the second order differential equation for $q$:
$$\frac{1}{B}\left(\frac{\rho \ddot{q}}{S} +\frac{\dot{q}}{S.e}\right) +\frac{B.d.\dot{q}}{m} =g$$
$$\ddot{q} +\left(\frac{B}{\rho e} +\frac{d.S.B^{2}}{\rho m}\right)\dot{q} =\frac{gBS}{\rho}$$
With the initial condition $\dot{q}_{(}{}_{0}{}_{)} =0$, we have the function $\dot{q}_{(}{}_{t}{}_{)}$:
$$\dot{q} =\frac{gBS}{\rho \alpha }\left( 1-e^{-\alpha t}\right)$$,in which $$\displaystyle \alpha =\frac{B}{\rho e} +\frac{d.S.B^{2}}{\rho m}$$
Take the intergral, we have the function $q$ of $t$:
$$q=\frac{gBS}{\rho \alpha }\left( t-\frac{1}{\alpha } +\frac{1}{\alpha } e^{-\alpha t}\right)$$
Then we have $v$:
$$\displaystyle v=\frac{1}{B}\left(\frac{\rho \dot{q}}{S} +\frac{q}{Se}\right) =\frac{g}{\alpha }\left( 1-e^{-\alpha t}\right) +\frac{g}{\rho e \alpha }\left( t-\frac{1}{\alpha } +\frac{1}{\alpha } e^{-\alpha t}\right)$$
So I think after a long time of falling, the disk won't reach a terminal velocity. Instead, it will experience a constant acceleration of $\frac{g}{\rho e_{0} \alpha }$
$e$ is the electric constant, I don't know why it cannot be displayed

Is it correct?

Sorry for any grammar or spelling mistakes, English isn't my first language T.T

 

[TSny]

From the given assumptions and the second Newton law, we can set up two equations:
$$\dot{q} =\frac{S}{\rho }\left( v.B-\frac{q}{S.e}\right)$$ (1)
$$\dot{v} =g-\frac{B.d.\dot{q}}{m}$$ (2)

Looks good.

Take the derivative of (1) over time, then substitute to (2), we have the second order differential equation for $q$:
$$\frac{1}{B}\left(\frac{\rho \ddot{q}}{S} +\frac{\dot{q}}{S.e}\right) +\frac{B.d.\dot{q}}{m} =g$$
$$\ddot{q} +\left(\frac{B}{\rho e} +\frac{d.S.B^{2}}{\rho m}\right)\dot{q} =\frac{gBS}{\rho}$$

Check the last equation above. In the parentheses on the left side, the first term should not have $B$. It should be $\dfrac{1}{\rho \epsilon_0}$.

This will change your expression for $\alpha$ where you have

$$\displaystyle \alpha =\frac{B}{\rho e} +\frac{d.S.B^{2}}{\rho m}$$

It should be
$$\displaystyle \alpha =\frac{1}{\rho \epsilon_0} +\frac{d.S.B^{2}}{\rho m}$$
Note that $S d$ is the volume of the disk. So $\alpha$ can be expressed in terms of the volume mass density $\mu$ of the material of the disk: $$\displaystyle \alpha =\frac{1}{\rho \epsilon_0}+\frac{B^2}{\rho \mu} = \frac{1}{\rho \epsilon_0}\left(1+\frac{B^2 \epsilon_0}{\mu}\right)$$ Since $\epsilon_0$ has the value 8.85 x 10^-12 C² /(Nm²), the dimensionless quantity $\dfrac{B^2 \epsilon_0}{\mu}$ inside the parentheses will be negligible compared to 1 for any reasonable values of B and $\mu$.

For the acceleration of the disk I find $$ a(t) = \frac{g}{1+\frac{B^2 \epsilon_0}{\mu}} \left(1+ \frac{B^2 \epsilon_0}{\mu}e^{-\alpha t}\right).$$ Since $\dfrac{B^2 \epsilon_0}{\mu} \ll 1$ for any reasonable choices of $B$ and $\mu$, we see that $a(t) \approx g$ to a very good approximation.

 

[AmanWithoutAscarf]

Wow, thank you so much!

Looks good.


Check the last equation above. In the parentheses on the left side, the first term should not have $B$. It should be $\dfrac{1}{\rho \epsilon_0}$.

This will change your expression for $\alpha$ where you have

It should be
$$\displaystyle \alpha =\frac{1}{\rho \epsilon_0} +\frac{d.S.B^{2}}{\rho m}$$
Note that $S d$ is the volume of the disk. So $\alpha$ can be expressed in terms of the volume mass density $\mu$ of the material of the disk: $$\displaystyle \alpha =\frac{1}{\rho \epsilon_0}+\frac{B^2}{\rho \mu} = \frac{1}{\rho \epsilon_0}\left(1+\frac{B^2 \epsilon_0}{\mu}\right)$$ Since $\epsilon_0$ has the value 8.85 x 10^-12 C² /(Nm²), the dimensionless quantity $\dfrac{B^2 \epsilon_0}{\mu}$ inside the parentheses will be negligible compared to 1 for any reasonable values of B and $\mu$.

For the acceleration of the disk I find $$ a(t) = \frac{g}{1+\frac{B^2 \epsilon_0}{\mu}} \left(1+ \frac{B^2 \epsilon_0}{\mu}e^{-\alpha t}\right).$$ Since $\dfrac{B^2 \epsilon_0}{\mu} \ll 1$ for any reasonable choices of $B$ and $\mu$, we see that $a(t) \approx g$ to a very good approximation.