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?