A light charged particle is revolving in a circle of radius r in electrostatic attraction of a static heavy particle with opposite charge. How does the magnetic field B at the centre of the circle due to the moving charge depend on r?

Options

A. `Bprop(1)/(r)`
B. `Bprop(1)/(r^(2))`
C. `Bprop(1)/(r^(3//2))`
D. `Bprop(1)/(r^(5//2))`

Explanation

Let a light particle of mass m and charge `-q_(2)` revolve around a heavy particle of charge `+q_(1)` in a circular path of radius r with a velocity v
Now, `(mv^(2))/(r)=(1)/(4piepsi_(0))((+q_(1))xx(-q_(2)))/(r^(2))`
or, `v^(2)=(1)/(4piepsi_(0)).(-q_(1)q_(2))/(mr)`,
`:.vprop(1)/(sqrtr)`

Time period of the charge in circular orbit,
`T=(2pir)/(v)`
`:.` Current due to the revolving charge,
`I=(-q_(2))/(T)=-(q_(2)xxv)/(2pir)`
`:.Iprop(v)/(r)or,Iprop(1)/(r^(3//2))[:'vprop(1)/(sqrtr)]`
So, magnetic field at the centre of the circle due to the moving charge `+q_(2)`,
`B=(mu_(0)I)/(2r)or,Bprop(I)/(r)`
or, `Bprop(1)/(r^(3//2)xxr)or,Bprop(1)/(r^(5//2))`.