Magnetic field...問左好多人都唔識..

chung

2008-03-17 5:38 am

A charge Q is smeared uniform over a thin plastic disk of radius R.It is then
spun about its axis at a apeed of N revoulations per second.

(a) Find the magnetic field produced by the charge at the center of the disk.
(b)What current at the rim of disk would produce the same magnetic field if
the disk were held stationary ?

回答 (1)

天同

2008-03-17 7:05 am

✔ 最佳答案

The surface density of charge s = (Q/pi.R^2), where pi=3.14159....

Consider a ring of radius r from the centre with width dr,
the charge on this ring = s.(2.pi.r).dr

Since the disk is rotating at N rev/s, the amount of charge dq passing through a reference point on the ring in 1 second is dq = [s.(2.pi.r).dr].N
thus, the equivalent current dI = dq/dt = [s.(2.pi.r).dr].N

the magnetic field at the centre of the disk dB due to this current is
dB = u.dI/2r
where u is the permeability of free space.

thus dB = u [s.(2.pi.r).dr].N /2.r = u.s.pi.N.dr
just integrate both sides with respect to r from r=0 to r=R to find the total B value.

(b) Having found B in part (a)
B = u.I/2R
i.e. I = 2R.B/u
calculate I

👍 0 👎 0