Magnetic field of revolving disk

[Dell]

find the expression for the magnetic field at a distance z on the axis which passes through the centre of a disk, with a radius of R, and a charge density σ, the disk is rotating at an anular velocity of ω.

http://lh4.ggpht.com/_H4Iz7SmBrbk/SivHVn0m_fI/AAAAAAAABC0/9sgJmbkoJK8/s720/Untitled.jpg

could someone tell me if this is all done correctly?

the equations i am going to use are
1)F=ILxB
2)f=qvxB
3)biot savar

1st stage i want to express the charge q,
q=σπR²

now i use
f=qvxB=σπR²vB ( the angle is 90 degrees constant)
f=σπR³Bω

F is also ILB so i know that IL=σπR³ω

now using biot savar

dB=(μI*dlxr)/4πr³)

now for I*dl i want to substitute the expression i found earlier, with R being my variable, not that sure about this...

dB==(μσπR³ωdR)/4π((R²+z²)^0.5)3)

dB=(μσR³ωdR)/4((R²+z²)^1.5)

B=(μσω/4)*∫R³dR/(z²+R²)^1.5

after integration i get

B=(μσω/4)*(1+z²/(z²+R₂))*(z²+R₂)^0.5 with my limits being R from 0 to R

and eventually i get
B=(μσω/4)*[(1+z²/(z²+R₂))*(z²+R₂)^0.5-2z]

i have never solved anything of this sort and hope everything i have done is okay, thanks

 

[Cyosis]

You're pretty close, however your mistake is to assume that the force is constant for every r which it is not. $F=qvB$, but v and q are different at different radii. Therefore the total force would be [itex]\int_0^{2R} vBdq =\int_0^{2R} \omega r B 2 \pi r \sigma dr=\int_0^{2R} I B dr[/tex]. We're integrating over the same range and variable so the integrands need to be the same therefore.<br /> <br /> $$2 \pi \sigma \omega r^2 B dr=I B dr \Rightarrow I dr= 2 \pi \sigma \omega r^2 dr$$<br /> <br /> Then using Biot-Savart yields:<br /> $$\frac{\mu_0}{4 \pi} \int_0^{2R} \frac{2 \pi \sigma \omega r^2}{r'^2} \cos \theta dr$$<br /> <br /> r' is the distance from a point on the disk to the z-axis, $r'^2=r^2+z^2$. And $\theta$ is the angle between r and r', so $\cos \theta$ picks out the x component of the magnetic field.<br /> <br /> An easier way would be using the surface current density, $K=\sigma v$. Biot-Savart then becomes:<br /> $$\vec{B}(\vec{r})=\frac{\mu_0}{4 \pi} \int \frac{\vec{K} \times \hat{r}}{r^2}da'$$[/itex]