What is the magnetic induction at the centre of a disc
Answers
Answer:
Here, a non-conducting thin disc of radius R charged uniformly over one side with surface density \sigmaσ rotates about its axis with an angular velocity \omegaω.
Let, qq be the uniform charge over one side of disc.
The current induced on the small area of disc having charge dqdq in time dtdt is
dI = \dfrac{dq}{dt}dI=
dt
dq
.......................................................(1)
\sigma = \dfrac{dq}{ds}σ=
ds
dq
\Rightarrow dq = \sigma ds⇒dq=σds
Also, dt = \dfrac{1}{n}dt=
n
1
and \omega = 2\pi n \Rightarrow n = \dfrac{\omega}{2\pi}ω=2πn⇒ n=
2π
ω
Using above equations we can rewrite equation (1) as
dI = \dfrac{\sigma ds \omega}{2\pi}dI=
2π
σdsω
Integrating both sides we get,
I = \sigma \omega \dfrac{2\pi R}{2\pi}I=σω
2π
2πR
I = \sigma \omega RI= σωR.................................................. (2)
According to Biot-Savart's law, the magnetic induction at the center of the disc due to small length dldl is
B = \int dB = \int \dfrac{\mu_0 I}{4\pi R}dlB=∫ dB=∫
4πR
μ
0
I
dl
B = \dfrac{\mu_0}{4\pi} \int (\sigma \omega R)\dfrac{dl \sin \theta}{R^2}B=
4π
μ
0
∫(σωR)
R
2
dlsinθ
....................................................................from (2)
Integrating both sides we get
B = \dfrac{\sigma \omega \mu_0 }{4\pi }\int \dfrac{(R)(2\pi R) \sin 90^\circ}{R^2} dRB=
4π
σω μ
0
∫
R
2
(R)(2πR)sin90
∘
dR..............((since dldl is almost straight, \theta = 90^\circ)θ= 90
∘
)
B = \dfrac{\sigma \omega \mu_0 }{2}\int dRB=
2
σω μ
0
∫dR
B = \dfrac{\mu_0 (\sigma \omega R)}{2}B=
2
μ
0
(σωR)