A non-conducting thin disc of radius R rotates about its axis with an angular velocity `omega`. The surface charge density on the disc varies with the distance r from the centre as `sigma(r)=sigma_(0)[1+((r)/(R))^(beta)]`, where `sigma_(0)` and `beta` are constants. If the magnetic induction at the center is `B=((9)/(10))mu_(0)sigma_(0)omegaR`, the value of `beta` is

Options

A. `(1)/(4)`
B. 4
C. `(1)/(2)`
D. 2

Answer

A

Explanation

Total charge on the disc, `Q=sigma(piR^(2))`
Period of motion, `T=2pi//omega`

So, current in the ring, `dI=(dQ)/(T)`
`because" "Q=sigma pi R^(2)impliesdQ=2sigma pi rdr`
`:." "dI=(2sigma pi rdr)/(2pi)xx omega=sigma" rdr"`
Magnetic fiedl due to ring is
`dB=(mu_(0)dI)/(2r)`
So, magnetic field due to disk is
`B=int dB=int_(0)^(R)(mu_(0))/(2r).sigma omega" rdr"`
On putting value of `omega`, we get
`B=(mu_(0)sigma_(0)omega)/(2)int_(0)^( R )[1+((r)/(R))^(beta)]dr`
`=(mu_(0)sigma_(0)omega)/(2)[R+(R^(beta+1))/(R^(beta)(1+beta))]=(mu_(0)sigma_(0)omega)/(2)xx((2+beta)/(1+beta))R`
Now,
`B=(9)/(10)mu_(0)sigma_(0)omegaR=(mu_(0)sigma_(0)omegaR)/(2)xx((2+beta))/((1+beta))`
`implies" "5beta+10=9beta+9`
`implies" "beta=(1)/(4)`