A surface charge density of `10^(-2)C//m^2` has been created by friction on a glass disk of h =5 mm thickness and R= 50 mm radius. The disk rotates at 1.6 r.p.s. Find the magnetic field intensity at the centre of the disk.
When you have learned to integrate, find the magnetic moment and the ratio of the magnetic moment to the angular momentum.

Answer

`2.5 xx 10^(-3)A//m; 4.9 xx 10^(-7) m A//m^(2) 4.1 xx 106(-4) C//kg`

Explanation

Consider the disk to be divided into thin concentric rings. The area of the ring is `triangleq =sigma triangle S=2pi sigma triangler`. When rotating the charge creates a magnetic field at the centre of the ring wih strength `triangleH=(triangleq)/(4pi r^2)=(sigma omega triangle r)/2`
The total intensity of the field at the centre is
`H=triangleH_(1)+triangleH_(2)+.....=1/2 sigma omega (triangler_1+triangler_2+.....)=1/2 sigma omega R`
where R is the external radius of the disk.
The magnetic moment of the ring is
`triangleP_(m)=ipir^2=pir^2 =(triangleq)/T=1/2 triangleq wr^2=pi sigma omegar^2 triangler`
To find the total magnetic moment of the rotating disk we must add up all these values. We have
`Pm=underset(0)overset(R)int pi sigma r^3dr=1/4 pi sigma omega R^4`
The moment of momentum `L=I omega=1/2 mR^2w=1/2pi wphR^4`, where p is the density of the material. We have `p_m/L=sigma/(2ph)`