# Total current with nonuniform current density

1. Dec 4, 2012

### hansbahia

1. The problem statement, all variables and given/known data
J=J(r)(θhat), J(r)=(3ITr2)/R3

a)total current of the disk?
b)J(r) was brought by a charge disk spinning at ang vel. ω, find the surface charge dens σ(r)?

2. Relevant equations
J=I/πr2
I=∫Jda
J=vσ
3. The attempt at a solution
a)I assumed I would have to integrate J(r)da from 0 to R to get the total current

I=from 0 to R∫Jda=from 0 to R∫(3ITr2)/R32πrdr
I=(3πITR)/2
But the result doesn't make sense?
How can I prove?

b) σ= (3ITr2(θhat))/R3ωrθ
am I using the right expression "θ" for the rotating angle or should I use ∅?

Last edited: Dec 4, 2012
2. Dec 4, 2012

### TSny

Note the direction of the current density. Draw a circle representing the disk and draw a radius of the circle. For several points along that radius, draw a vector representing the current density.

Note also that the given current density function J(r) has units of A/m rather than A/m2. So you are dealing with a current density that specifies current per unit length rather than current per unit area.

3. Dec 4, 2012

### hansbahia

The direction would be counterclockwise

Okay, how would that affect my calculation?
I'm sorry if I confused you by writing big J instead of small j. Here J is not current per unit area

Last edited: Dec 4, 2012
4. Dec 4, 2012

### TSny

OK, so the current is circling CCW around the center of the disk. If you draw a radius of the disk, then the current crosses perpendicularly to the radius. If you go out the radius a distance r and mark off an infinitesimal increment in radius, dr, how would you express the amount of current through dr?

5. Dec 4, 2012

### hansbahia

.._
|→ | → →→ I .....|
.\_/

dI=j(r)dr? j(r) would be in current density per unit-length, or J(r) like the question

Sorry I attempted to draw a circle with current going perpendicular to R

Last edited: Dec 4, 2012
6. Dec 4, 2012

### TSny

Good. So calculate the total current crossing the entire radius R.

7. Dec 4, 2012

### hansbahia

dI=j(r)dr
I= ∫j(r)dr
I=IT

Last edited: Dec 4, 2012
8. Dec 4, 2012

9. Dec 4, 2012

### TSny

Now the current is due to spinning of a surface charge density σ(r) on the disk at angular speed ω.

Go out along the radius a distance r and again pick an interval dr. Assume this radial line does not spin with the disk but remains fixed in the laboratory frame. How much charge per second passes through dr due to the spinning?

10. Dec 4, 2012

### hansbahia

dI=dQ/T=(2πσrdr)/(2π/ω)=ωσrdr

σ=j(r)/ωr

Last edited: Dec 4, 2012
11. Dec 4, 2012

### TSny

Your result for ω looks good. Not sure why you want an angle in there. Note that ω is angle per second. Don't know if that helps.

12. Dec 4, 2012

Thank you!