Does a charged insulating ring turn if a B-field is switched on?

[phantomvommand]

Homework Statement

Charge Q is uniformly distributed on a thin *insulating* ring of mass m, which is initially at rest. To what angular velocity will the ring be accelerated when a magnetic field B, perpendicular to the plane of the ring, is switched on?

Relevant Equations

Integrate E . ds= - dΦ/dt
dF = E dq
dT = r dF, T is torque, r is radius of ring
T = Iα
a = rα

The physics behind this problem is that an electric field is induced (by Faraday's Law), when the B field is switched on. Charges on the ring now experience a force as given by dF = E dq. Apparently, because of this, the ring starts rotating.

I understand that charges in an insulating material are not allowed to move; thus, if they experience a force, does it mean the entire material moves with them?

Thank you!

 

[phyzguy]

I understand that charges in an insulating material are not allowed to move; thus, if they experience a force, does it mean the entire material moves with them?

Yes. The entire ring will start to rotate. You are probably intended to assume that the ring is a rigid body.

 

[etotheipi]

Taking a curve $\Gamma = \partial S$ around the ring you can write down$$\oint_{\Gamma} \mathbf{E} \cdot \mathrm{d}\boldsymbol{x} = - \frac{\mathrm{d}}{\mathrm{d}t} \int_S \mathbf{B} \cdot \mathrm{d}\mathbf{S}$$I think you might need to assume something about how specifically the magnetic field is "switched on"; for instance, suppose that it increases linearly at a rate $\lambda$ to its final value $\mathbf{B}_0$ at a time $\frac{1}{\lambda}$, i.e. that $\mathbf{B}(\boldsymbol{x}, t) = \lambda t \mathbf{B}_0$. With that you should be able to calculate something for $- \frac{\mathrm{d}}{\mathrm{d}t} \int_S \mathbf{B} \cdot \mathrm{d}\mathbf{S} = - \frac{\mathrm{d}}{\mathrm{d}t} \left( \mathbf{B} \cdot \mathbf{A} \right)$. With any luck the $\lambda$ will drop out, so you can imagine making it arbitrarily high.

As for the other side of the equation, I think it might be easiest to see that the moment $\mathrm{d}M$ on a little element $\mathrm{d}q = \rho |d\boldsymbol{x}|$ is going to be the radius $R$ multiplied by the component of the force perpendicular to the radius, $\mathrm{d}q \mathbf{E} \cdot \hat{\mathbf{t}} = \rho \mathbf{E} \cdot \mathrm{d}\boldsymbol{x}$, or in other words$$\mathrm{d}M = \rho R (\mathbf{E} \cdot \mathrm{d}\boldsymbol{x})$$How does that look to you? Maybe try going from there.

 

[kuruman]

I think you might need to assume something about how specifically the magnetic field is "switched on"; for instance, suppose that it increases linearly at a rate $\lambda$ to its final value $\mathbf{B}_0$ at a time $\frac{1}{\lambda}$, i.e. that $\mathbf{B}(\boldsymbol{x}, t) = \lambda t \mathbf{B}_0$. With that you should be able to calculate something for $- \frac{\mathrm{d}}{\mathrm{d}t} \int_S \mathbf{B} \cdot \mathrm{d}\mathbf{S} = - \frac{\mathrm{d}}{\mathrm{d}t} \left( \mathbf{B} \cdot \mathbf{A} \right)$. With any luck the $\lambda$ will drop out, so you can imagine making it arbitrarily high.

The details about how the field increases are irrelevant. That's because the total torque on the ring is $\tau=\dfrac{dL}{dt}$ ($L$ is the angular momentum) and there is a term $\dfrac{dB}{dt}$ on the right hand side of the last equation posted in #3 by @etotheipi. There are exact differentials on both sides of the equation which means that $\Delta L$ is $\Delta B$ times a constant.

 

[Gordianus]

I can't help quoting Feynman (sort of):"Where does the angular momentum come from?"

 

[kuruman]

I can't help quoting Feynman (sort of):"Where does the angular momentum come from?"

The same place it goes back to when you turn the field off?

 

[Gordianus]

The same place it goes back to when you turn the field off?

Excellent! I hadn't realized that switching the field on and off deals with that pesky question.

 

[etotheipi]

The details about how the field increases are irrelevant. That's because the total torque on the ring is $\tau=\dfrac{dL}{dt}$ ($L$ is the angular momentum) and there is a term $\dfrac{dB}{dt}$ on the right hand side of the last equation posted in #3 by @etotheipi. There are exact differentials on both sides of the equation which means that $\Delta L$ is $\Delta B$ times a constant.

Okay that makes sense - it's satisfying to know! In that case, I guess not dissimilar to working with state functions in thermodynamics; we should just pick the simplest possible 'switching on behaviour' and work with that, which I think is probably the linear variation.

 

[kuruman]

Okay that makes sense - it's satisfying to know! In that case, I guess not dissimilar to working with state functions in thermodynamics; we should just pick the simplest possible 'switching on behaviour' and work with that, which I think is probably the linear variation.

Perhaps you misunderstood my point, perhaps I misunderstood yours. You don't need a "switching on behaviour" to figure out the angular momentum picked up by the ring. You can write $L[B(t)]$ so if you know $B$, you can find $L$ regardless of $B(t)$.

Yes, it is akin to a state function. The same applies to the total charge that would flow in a wire loop if you switch the field ON/OFF or even if you twist the wire loop to a pretzel shape while the field varies in time. That's because you have $\frac{dQ}{dt}$ on one side of the equation and $\frac{d\Phi_M}{dt}$ on the other. Assuming constant loop resistance, the net charge that flows, $\Delta Q=\int I~dt$, is proportional to the net change in flux, $\Delta \Phi_M$, because $dQ$ is proportional to $d\Phi_M$ with the same proportionality constant at all times.

 

[etotheipi]

That sounds okay - forgive me, if I'm missing your point. Let's just do it and see... from where I left off we have (denoting $L \equiv L_z$, $B \equiv B_z$, $M \equiv M_z$)$$\frac{1}{\rho R} \int \mathrm{d}M = \frac{1}{\rho R} \frac{\mathrm{d}L}{\mathrm{d}t} = - A \frac{\mathrm{d}B}{\mathrm{d}t}$$or in other words$$-\frac{1}{\rho R A} \int_0^t \frac{\mathrm{d}L}{\mathrm{d}\xi} \mathrm{d}\xi = \int_0^{t} \frac{\mathrm{d}B}{\mathrm{d}\xi} \mathrm{d}\xi \implies - \frac{L(t)}{\rho R A} = - \frac{I \omega(t)}{\rho R A} = B(t)$$so for whatever $\mathbf{B}: \mathbb{R} \rightarrow \mathbb{R}^3$ you choose, at a given time $t$ the angular momentum $\mathbf{L}(t)$ is just given by the above, like you mentioned in post #4. That's the sense in which the final angular momentum is analogous to a state function. So we can just arbitrarily imagine the case when the 'switching on behaviour' consists of the magnetic field increasing linearly with time, or indeed any other more complicated behaviour?

Sorry if I still misunderstood, it's been a long day haha

 

[etotheipi]

Oh hang on, that was your point wasn't it? That you don't actually need to think about how the magnetic field varies when it's being switched on.

Yes, okay I feel slightly stupid now. Basically, just ignore everything I said above