An electric current I enters and leaves a uniform circular wire of radius a through diametrically opposite points. A charged paricle q moving along the axis of the circular wire passes through its centre at speed v. The magnetic force acting on the particle when it passes through the centre has a magnitude

Options

A. `qv(mu_(0)i)/(a)`
B. `qv (mu_(0)i)/(2a)`
C. `qv (mu_(0)i)/(2pi a)`
D. zero

Answer

D

Explanation

From Amper.s circuital law, we get `oint vec(B).d vec(l)= mu_(0) I_("enc")` Now, at the centre, `I_("enc")=0`, thus `oint vec(B).d vec(l)=0`
`rArr vec(B)=0 [ :. d vec(l)=0]`
`:.` Magnetic force experienced by the particle when it passes through the centre of the circle is
`vec(F)= q (vec(v) xx vec(B))=0`