# Magnetic Field of a thin disk

1. Dec 1, 2003

### discoverer02

I'm not quite getting the following problem:

A thin disk of dielectric material, having a total charge +Q uniformly distributed over its surface, and having radius a, rotates n times per second about an axis through the center of the disk and perpendicular to the disk.

Show that the magnetic field produced at the center of the disk is unQ/a.

I know that B at the center of a wire ring is uI/2r. If I start here then dB = udI/(2dr).

I = dq/dt. In this case I = nQ/sec, so can the equation become
db = undQ/(2dr)? I'm sure this isn't right but here goes.

Q/(pi)a^2 = dq/2(pi)rdr so dq = 2Qrdr/a^2 ?

If I plug this into my equation the dr's cancel, so I must be doing something wrong?

Thanks in advance for the help.

2. Dec 1, 2003

### Staff: Mentor

Hint: Find the current for a rotating ring of charge of width dr. (Find the charge per unit area; from that find the charge in each ring.)

3. Dec 1, 2003

### HallsofIvy

Staff Emeritus
Actually, your original equations don't quite make sense:
dB= &mu;dI/(2dr)has a differential on one side and a derivative on the other. Also, I don't believe I= dq/dt= nQ/sec. Since the disk is rotating at n revolutions per second, it is rotating at 2&pi;n radians per second and a piece at distance r (linear units) from the center is moving at 2&pi;nr linear units per second. A thin ring, distance r from the center and width dr would have area 2&pi;rdr and so charge (2&pi;rdr)(Q/(&pia2)= 2rQdr/a2.
I= dq/dt= 2&pi;nr(2rQdr/a2)= (4&pi;Qnr2/a2)dr.
That makes dB= &mu;I/(2r)= 2&pi;&mu;Qnr/a2 (this is "dB" because it is a small part of the whole magnetic field due to a small part of the entire radius. To find B, integrat from r= 0 to r= a.

4. Dec 1, 2003

### discoverer02

OK, let's see if this works.

Charge per unit area for the whole disk is Q/(pia^2).
The ring's charge per unit aread would be dQ/2pirdr.

Therefore dQ = 2Qrdr/(a^2).

Since every second ndQ charges flow through what could be considered a cross sectional area, can I substitute ndQ for I?

If I can then dB = nu2Qrdr/(2ra^2) = nuQdr/a^2.

If I integrate this from 0 to a, I get the correct answer:
uNQ/a.

But is the math and reasoning OK or did I force something that I shouldn't have.