An alpha particle moves along a circular path of radius `2Å` with a uniform speed of `2xx10^6ms^-1`. Calculate the magnetic field set up at the center of circular path.

Answer

`1*6T`

Explanation

Charge on `alpha`-particle, `q=+2e`, where e is the charge of an electron. Thus
`q=2xx1*6xx10^(-19)C`.
`r=2xx10^(-10)m`, `v=2xx10^6ms^-1`
Current `I=q/T=(q)/(2pir//v)=(qv)/(2pir)`
Magnetic field at the centre of circular path
`B=(mu_0)/(4pi)(2piI)/(r)=(mu_0I)/(2r)=(mu_0)/(2r)((qv)/(2pir))=(mu_0)/(4pi)(qv)/(r^2)`
`=(10^-7xx(2xx1*6xx10^(-19))xx(2xx10^6))/((2xx10^(-10))^2)=1*6T`