db = μidl/4πr2
let the angular velocity we ω
Now consider a small area at distance x from the centre = 2Ï€xdx
Q=σ2πxdx
i = ωσxdx
db = μ*ωσxdx*2πx/4πx2
b = μωrσ/2
This is a simple standard question... (Which I would rate as a must know for most beginners)
A disc of uniform surface charge density σ and radius r is rotated about its center, find the magnetic field induced at the center of the disc.
db = μidl/4πr2
let the angular velocity we ω
Now consider a small area at distance x from the centre = 2Ï€xdx
Q=σ2πxdx
i = ωσxdx
db = μ*ωσxdx*2πx/4πx2
b = μωrσ/2
Awesome work virang...
:)
Just elaborate how you found out current.. so that it becomes a bit more lucid for the other users..
Current = Q * no.of times it passes a area
i = Q*Revs
i = Q*ω/2π
Q = σ * 2πxdx
Therefore i = σωxdx