A charge q is accelerated through a potential difference V . It is then passed normally through a uniform magnetic field, where it moves in a circle of radius r. Then potential difference required to move it in a circle of radius 2 r is

Options

A. 2 V
B. 4 V
C. 1 V
D. 3 V

Answer

B

Explanation

If v is the velocity acquired by the charge when accelerated through the potential difference V, then
`1/2mv^2=qV " or "v=sqrt((2qV)/m)`
When it passes normally through of uniform magnetic field B, it moves in the circle of radius r.
`:.(mv^2)/r=qvB`
or `r=(mv)/(qB)=m/(qB)sqrt((2qV)/m)=1/Bsqrt((2mV)/q)`
or `V=(B^2qr^2)/(2m)" "...(i)`
Let V' be the required potential difference in order to move it in the circle of 2r.
`V'=(B^2q(2r)^2)/(2m)=4((B^2qr^2)/(2m))=4V`