Terminal velocity of a falling conducting ring

So you would have to consider both the magnetic and electric forces acting on the loop to find the net force on it.
  • #1
quasar_4
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Homework Statement



A conducting ring with radius a and mass m is placed in a magnetic field at a height H above the origin of the reference frame. The plane of the ring is parallel to the ground, that is, the normal is directed along the z-axis. The electrical resistance per unit length of the ring is R/(2Pia), so that the resistance of the loop is R. The magnetic field has spatial dependence:

[tex]\vec{B} = \frac{B_0}{L}\left(-\frac{\rho}{2} \hat{\rho} + z \hat{z} \right)[/tex]

where [tex] \hat{\rho}, \hat{z} [/tex] are unit vectors in the usual cylindrical coordinate system.

At a certain time, the ring is dropped and falls due to gravity. The plane of the ring remains horizontal as the ring falls. Find the terminal velocity of the ring.

Homework Equations



We've got Maxwell's equations, F=ma, Lorentz force F=qv X B, etc.

The Attempt at a Solution



Here's my attempt so far:

[tex] \nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} \Rightarrow \oint{\vec{E} \cdot d\vec{l}} = -\frac{d}{dt} \int{\vec{B} \cdot \hat{n} \hspace{1mm} da} \Rightarrow E(2 \pi a) = -\frac{d}{dt} \int{B_z \hspace{1mm}da} \Rightarrow E(2 \pi a) = -\frac{d}{dt} \int_0^{2\pi}{d\phi}\int_0^a{B_z \rho\hspace{1mm} d\rho} = - \frac{B_0 a^2 \pi}{L}\frac{dz}{dt} [/tex]

But dz/dt is just the velocity we're looking for!

Here is where I am now stuck. I have a nice expression for the electric field, E, in terms of velocity, dz/dt =v:

[tex] \vec{E} = - \frac{B_0 \pi a v_z}{2L} \hat{\phi} [/tex]

But the electric field points in the circumferential direction, not the z direction. In terms of finding terminal velocity, I want something to the effect of

[tex] \sum F_z = -mg + F_? = ma [/tex], where F? is the net force in the z direction due to my induced electric field. From this point it is easy to set ma =0 and solve for v-terminal. But I can't just say something like F=qE, because E points in the wrong direction.

Furthermore, I haven't at all used the fact that the ring is at a height H or that it has resistance R - are those unnecessary bits of information, or am I approaching this entirely wrong? I almost could solve this using conservation of energy, but it's the terminal velocity I want and not clear to me how to find that in an energy formulation of the problem (might be possible, though). So - can anyone help? o:)
 
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  • #2
Oh, one more note. I also thought of this:

[tex] -\frac{d}{dt} \int B \cdot da = -\frac{d \mathcal{E}}{dt} [/tex]

, i.e. the change in magnetic flux. And then we could use the fact that

[tex] \vec{F}_{mag} = I \int d\vec{l} \times \vec{B} = \frac{\mathcal{E}}{R} \int d\vec{l} \times \vec{B} [/tex]

but then when I do out the cross product I get for the z-component of magnetic force

[tex] \vec{F}_{mag} = \frac{B_0 \rho \pi a}{L} \hat{z} [/tex]

which has no expression for velocity in it. So, how would I go about finding v-terminal from this?
 
  • #3
quasar_4 said:
[tex] \nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} \Rightarrow \oint{\vec{E} \cdot d\vec{l}} = -\frac{d}{dt} \int{\vec{B} \cdot \hat{n} \hspace{1mm} da} \Rightarrow E(2 \pi a) = -\frac{d}{dt} \int{B_z \hspace{1mm}da} \Rightarrow E(2 \pi a) = -\frac{d}{dt} \int_0^{2\pi}{d\phi}\int_0^a{B_z \rho\hspace{1mm} d\rho} = - \frac{B_0 a^2 \pi}{L}\frac{dz}{dt} [/tex]

You seem to be assuming (without any justification) that [itex]\oint\textbf{E}\cdot d\textbf{l}=2\pi aE[/itex]... is that really true for any old electric field?

quasar_4 said:
Oh, one more note. I also thought of this:

[tex] -\frac{d}{dt} \int B \cdot da = -\frac{d \mathcal{E}}{dt} [/tex]

, i.e. the change in magnetic flux. And then we could use the fact that

[tex] \vec{F}_{mag} = I \int d\vec{l} \times \vec{B} = \frac{\mathcal{E}}{R} \int d\vec{l} \times \vec{B} [/tex]

but then when I do out the cross product I get for the z-component of magnetic force

[tex] \vec{F}_{mag} = \frac{B_0 \rho \pi a}{L} \hat{z} [/tex]

which has no expression for velocity in it. So, how would I go about finding v-terminal from this?

Isn't there also an electric force on the loop?
 

FAQ: Terminal velocity of a falling conducting ring

1. What is the terminal velocity of a falling conducting ring?

The terminal velocity of a falling conducting ring is the maximum speed it will reach when falling through a fluid, such as air. It is determined by the balance of forces acting on the ring, including gravity and air resistance.

2. How is the terminal velocity of a falling conducting ring calculated?

The terminal velocity of a falling conducting ring can be calculated using the equation: vt = √(2mg/ρACd), where vt is the terminal velocity, m is the mass of the ring, g is the acceleration due to gravity, ρ is the density of the fluid, A is the cross-sectional area of the ring, and Cd is the drag coefficient.

3. Does the mass of the conducting ring affect its terminal velocity?

Yes, the mass of the conducting ring does affect its terminal velocity. The heavier the ring, the greater the force of gravity acting on it, and therefore, the higher the terminal velocity will be.

4. How does the shape of the conducting ring affect its terminal velocity?

The shape of the conducting ring can affect its terminal velocity through its cross-sectional area and drag coefficient. A larger cross-sectional area or a higher drag coefficient will result in a lower terminal velocity, as there is more air resistance acting on the ring.

5. Can the terminal velocity of a falling conducting ring change?

Yes, the terminal velocity of a falling conducting ring can change if there is a change in the forces acting on it. For example, if the density of the fluid or the shape of the ring changes, the terminal velocity will also change. Additionally, if the ring reaches a point where the forces are no longer balanced, such as when it reaches the ground, the terminal velocity will also change.

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