- #1
DrPapper
- 48
- 9
Homework Statement
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A very thin plastic ring (radius R) has a constant linear charge density, and total charge Q. The ring spins at angular velocity [itex] \omega [/itex] about its center (which is the origin). What is the current I, in terms of given quantities? What is the volume current density J in cylindrical coordinates? (This may be a little tricky, since the ring is "very thin", there will be some [itex] \delta [/itex] functions. I suggest writing down a formula for [itex] \rho [/itex](s, [itex] \phi[/itex], z) first. And, remember that J should be a vector!)
Homework Equations
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My problem is finding these, one might say it is "My real question".
The Attempt at a Solution
For the current I claimed that I=[itex]\omega \lambda[/itex] and [itex]\lambda [/itex]=Q/2[itex]\pi [/itex]R therefore I=[itex]\omega [/itex]Q/2[itex]\pi [/itex]R
I feel pretty strong about that on the basis that current is due to a moving charge and that the ring itself is a charge distribution that is spinning. However, if someone could verify that I'd be very grateful.
My real question is in regards to finding the delta function. My thoughts were that radially the ring is infinitely thin and it would also be going along the Z axis. However the length is just 2[itex]\pi[/itex]R so I just need to multiply these together as such my volume would become:
V=[itex]\delta[/itex](s-R)[itex]\delta[/itex](z-z')2[itex]\pi[/itex]R
I choose to use z' since the problem never actually tells us where along z the loop is situated.
Since [itex]\rho[/itex] is just charge per unit volume then:
[itex]\rho[/itex]=Q/[itex]\delta[/itex](s-R)[itex]\delta[/itex](z-z')2[itex]\pi[/itex]R
Once I have this figured out I can move on to finding J.