(a) Derive an expression for the force between two long parallel current carrying conductors .
(b) Use this expression to define S.I. unit of current .
(c) A long straight wire AB carries a current I.A proton P travels with a speed v , parallel to the wire , at a distance d from it in a direction opposite to the current as shown in the figure . What is the force experienced by the proton and what is its direction ?

Explanation

(c) Magnetic field induction at the point P is
`B = (mu_(0))/(4 pi) . (2I)/(r)`
The direction of `vecB` is perpendicular to the plane of paper directed inwards (According to Right Hand Thumb rule )
The force on moving proton of charge q due to magnetic filed B is
F = qvB sin `90^(@) = q v B `
Since charge of proton q = `1.6 xx 10^(-19) C`
`therefore F = 1.6 xx 10^(-9) xx v B N.`
The direction of force on proton , according to Flemming's left Hand Rule acts in the plane of paper towards right .