Electric current in a rotating ring

[Elysium]

I'm currently stuck on this question on the image attachment. Any help would be definitely ppreciated.

Ok, so from what I understand, it asks what is the current that passes through the fixed line.

For part (a), I see that the current is 'discontinuous', and I'm not enitrely sure how to solve it.

For part (b), I multiply the charge density $$\lambda = \frac{Q}{\pi a}$$ with the tangential speed of the ring $$a \omega$$. That would give me the charge over time, right? I believe I should of done this part with differentials though with a segment $$dQ = \lambda dr$$.

 

[Elysium]

*bump* still need help.

 

[topsquark]

I'm currently stuck on this question on the image attachment. Any help would be definitely ppreciated.

Ok, so from what I understand, it asks what is the current that passes through the fixed line.

For part (a), I see that the current is 'discontinuous', and I'm not enitrely sure how to solve it.

For part (b), I multiply the charge density $$\lambda = \frac{Q}{\pi a}$$ with the tangential speed of the ring $$a \omega$$. That would give me the charge over time, right? I believe I should of done this part with differentials though with a segment $$dQ = \lambda dr$$.

In the first problem you simply want to consider how much charge is passing through the indicated arc per unit time. In other words, don't use the derivative formula, use $$I= \Delta Q / \Delta t$$. The simplest would be to choose delta t as one period. How much charge passes through that arc in one period?

Your part b seems valid to me.

-Dan

 

[Elysium]

In the first problem you simply want to consider how much charge is passing through the indicated arc per unit time. In other words, don't use the derivative formula, use $$I= \Delta Q / \Delta t$$. The simplest would be to choose delta t as one period. How much charge passes through that arc in one period?

The full $$Q$$ of course, neglecting the bits on both poles that just spin.

So that would make $$Q / T$$ and

$$\omega = \frac{2 \pi}{T}$$
$$T = \frac{2 \pi}{\omega}$$

So that means the answer is:

$$I = \frac{Q \omega}{2 \pi}$$

Ok so that's the same answer as question (b). I guess that makes sense since they both have the same amount of Q passing through the same period. So what's the difference? One is done by substitution and the other by multiplying the density with the tangential speed?