Problem on induced electric field, electrodynamics

[phantomvommand]

Homework Statement

Please see the problem below

Relevant Equations

Emf = dΦ/dt
i = V/R = emf/R
Conservation of energy

The problem is shown above, the hint to solve the problem is below. See the hint if it is difficult for you to imagine what is going on.
I am assuming the diagram in the hint shows what's happening when the mass is falling at terminal velocity. I have quite a few questions.

1. How do the wheels start spinning? Does it exert a torque about the shaft (due to friction), which then spins the wheels?

2. How does the mass reach terminal velocity? What forces are acting to slow down the mass?

3. Why must the 2 wheels spin in opposite directions (clockwise and anticlockwise)? Is this due to the friction at the point of contact of the wheels?

4. How do I determine the direction of the induced electric field in each of the wheels?

Thanks to all for any help.

 

[etotheipi]

1. How do the wheels start spinning? Does it exert a torque about the shaft (due to friction), which then spins the wheels?

I think so, yes. The string connecting the hanging mass to the axle is in tension, so exerts a torque on the axle. Fundamentally this is indeed due to friction between the string and the axle, although the torque the string exerts about the axle comes out to simply be the tension $T$ multiplied by $a$.

(If you want to learn more, google the physics of a "capstan")

2. How does the mass reach terminal velocity? What forces are acting to slow down the mass?

The mass will reach terminal velocity here when the tension in the string becomes equal in magnitude to its weight. Although, as the hint suggests, thinking about forces and torques is perhaps not the most straight-forward approach to this problem!

3. Why must the 2 wheels spin in opposite directions (clockwise and anticlockwise)? Is this due to the friction at the point of contact of the wheels?

Pretty much, yes. The constraint is that there's no slippage between the wheels, so the velocities of the material on both wheels at that point must be equal (i.e. both upward, with the same speed). So the two rotate in opposite directions, with angular speeds inversely proportional to their radii.

4. How do I determine the direction of the induced electric field in each of the wheels?

I think the easiest way to explain this is a little bit-handwavey (!), but hopefully that's not too much of an issue. Take a wheel of radius $r$ and axle-radius $r_a$, rotating at $\Omega$, in a uniform magnetic field $B$. Consider a sector of angle $\delta \theta = \Omega \delta t$. You can prove by geometry that the area of this sector, minus the area of the little bit of the axle, is $$\delta A = \frac{\delta \theta}{2}(r^2 - r_a^2)$$Can you see how? What is the flux $\delta \phi$ through this element $\delta A$, and so what is $\delta \phi / \delta t$? Notice that the sign of $\mathcal{E} = - \delta \phi / \delta t$ depends linearly on $\Omega$, so whether the electric field is radially inward or outward depends on the direction of rotation.

If you know some vector calculus, you can do a bit of a better derivation by taking the line integral of $\mathbf{v} \times \mathbf{B} = (\boldsymbol{\omega} \times \boldsymbol{x}) \times \mathbf{B}$ along a radial path with line element $\mathrm{d}\boldsymbol{x} = \mathrm{d}s \hat{\boldsymbol{r}}$. That might make the signs a bit clearer!

Anyway, can you now see how to relate the change in the gravitational potential energy of the mass to the energy dissipated in the effective "resistor", in a time $\delta t$?

 

[phantomvommand]

I think the easiest way to explain this is a little bit-handwavey (!), but hopefully that's not too much of an issue. Take a wheel of radius $r$ and axle-radius $r_a$, rotating at $\Omega$, in a uniform magnetic field $B$. Consider a sector of angle $\delta \theta = \Omega \delta t$. You can prove by geometry that the area of this sector, minus the area of the little bit of the axle, is $$\delta A = \frac{\delta \theta}{2}(r^2 - r_a^2)$$Can you see how? What is the flux $\delta \phi$ through this element $\delta A$, and so what is $\delta \phi / \delta t$? Notice that the sign of $\mathcal{E} = - \delta \phi / \delta t$ depends linearly on $\Omega$, so whether the electric field is radially inward or outward depends on the direction of rotation.

If you know some vector calculus, you can do a bit of a better derivation by taking the line integral of $\mathbf{v} \times \mathbf{B} = (\boldsymbol{\omega} \times \boldsymbol{x}) \times \mathbf{B}$ along a radial path with line element $\mathrm{d}\boldsymbol{x} = \mathrm{d}s \hat{\boldsymbol{r}}$. That might make the signs a bit clearer!

Anyway, can you now see how to relate the change in the gravitational potential energy of the mass to the energy dissipated in the effective "resistor", in a time $\delta t$?

Thanks for the reply. I can see that the magnitude of emf induced is proportional to the magnitude of w, where w is the angular velocity of the plate. Why is E-field directed either radially inward or outward? Shouldn't E-field be in a circular direction, either clockwise or counterclockwise?

Does the direction of E-field go against the direction of w? In other words, if w is clockwise, would E-field be anticlockwise?

Thank you

 

[etotheipi]

Thanks for the reply. I can see that the magnitude of emf induced is proportional to the magnitude of w, where w is the angular velocity of the plate. Why is E-field directed either radially inward or outward? Shouldn't E-field be in a circular direction, either clockwise or counterclockwise?

Does the direction of E-field go against the direction of w? In other words, if w is clockwise, would E-field be anticlockwise?

It turns out that the electric field really is radial, although it's perhaps not obvious why. Although, one thing to note is that electric fields don't form closed loops like you're suggesting.

As a heuristic, imagine you have a thin charged rod hinged at one end, rotating at angular speed $\omega$. An element of charge $\delta q$ at radius $r$ has a speed $r\omega$ in the tangential anti-clockwise direction. If the magnetic field points into the page, say, then the magnetic force on this element points toward the origin.

However, because of the so-called "Hall effect", there'll also be an electric field set up in the rod. If we can take the mass of the element multiplied by its centripetal acceleration to be negligible compared to each of the electric and magnetic forces, then the electric force has to be exactly opposite to the magnetic force. In any case the electric field will point radially outwards.

To construct the disk, you can imagine partitioning the disk into $N$ sectors, and letting $N \rightarrow \infty$ in which case the sectors get narrower and more rod-like...

[On a separate note, I'm not sure if it totally makes sense to speak of the EMF along a path that is not closed, e.g. the radial one we were using above. Usually EMF is defined like$$\mathcal{E} = \oint (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \cdot d\mathbf{x}$$around a closed path. But here we used$$\mathcal{E}_{\text{motional}} = \int_a^b (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{x} = - \int_a^b \mathbf{E} \cdot d\mathbf{x}$$where the integration is along an open radial path. I'm going to call @TSny for help, to ask whether he can add some words of wisdom ]

 

[phantomvommand]

It turns out that the electric field really is radial, although it's perhaps not obvious why. Although, one thing to note is that electric fields don't form closed loops like you're suggesting.

As a heuristic, imagine you have a thin charged rod hinged at one end, rotating at angular speed $\omega$. An element of charge $\delta q$ at radius $r$ has a speed $r\omega$ in the tangential anti-clockwise direction. If the magnetic field points into the page, say, then the magnetic force on this element points toward the origin.

However, because of the so-called "Hall effect", there'll also be an electric field set up in the rod. If we can take the mass of the element multiplied by its centripetal acceleration to be negligible compared to each of the electric and magnetic forces, then the electric force has to be exactly opposite to the magnetic force. In any case the electric field will point radially outwards.

To construct the disk, you can imagine partitioning the disk into $N$ sectors, and letting $N \rightarrow \infty$ in which case the sectors get narrower and more rod-like...

[On a separate note, I'm not sure if it totally makes sense to speak of the EMF along a path that is not closed, e.g. the radial one we were using above. Usually EMF is defined like$$\mathcal{E} = \oint (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \cdot d\mathbf{x}$$around a closed path. But here we used$$\mathcal{E}_{\text{motional}} = \int_a^b (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{x} = - \int_a^b \mathbf{E} \cdot d\mathbf{x}$$where the integration is along an open radial path. I'm going to call @TSny for help, to ask whether he can add some words of wisdom ]

This has been very helpful, thank you very much!

 

[etotheipi]

Just to make things clearer, I thought I'd write the earlier suggestion up formally. Consider a disk lying in the $\mathrm{xy}$ plane, and let $\mathbf{B} = B \mathbf{e}_z$ and $\boldsymbol{\omega} = \omega \mathbf{e}_z$. The condition we identified on the fields (by neglecting the $m\mathbf{a}$ side of Newton's second law) was that$$\mathbf{E} + \mathbf{v} \times \mathbf{B} = 0$$The rigid body rotation satisfies $\mathbf{v} = \boldsymbol{\omega} \times \boldsymbol{x} = \omega \mathbf{e}_z \times r \mathbf{e}_r = \omega r \mathbf{e}_{\varphi}$, so $$\mathbf{v} \times \mathbf{B} = \omega r \mathbf{e}_{\varphi} \times B \mathbf{e}_z = \omega r B\mathbf{e}_r \implies \mathbf{E} = - \omega r B\mathbf{e}_r $$i.e. you have a radial electric field. Since we have $\partial \mathbf{B} / \partial t = 0$, the electric field is conservative and can be described by the gradient of a scalar potential $\phi$, as $\mathbf{E} = - \nabla \phi$. Hence you write$$\nabla \phi = \omega r B \mathbf{e}_r$$The cylindrical symmetry suggests an ansatz of $\phi = \phi(r)$, so you can substitute in $\nabla \phi = (\partial \phi / \partial r) \mathbf{e}_r$ which you can integrate to$$\phi(r) = \frac{1}{2} \omega B r^2$$just as before.