A long , straight wire carries a current i. A particle having a positive charge q and mass m, kept at a distance x_0 from the wire is projected towards it with a speed v. Find the minimum separation between the wire and the particle.

Explanation

Let the particle be initially at `P` (figure).Take the wire as the `Y`-axis and the foot of perpendicular from `P` to the wire as the origin.Take the line `OP` as the `X`-axis.We have,`OP=x_(0)`.The magnetic field `B` at any point to the right of the wire is along the positive `Z`-axis.The magnetic force on the particle is, therefore,in the `X-Y` plane.As there is no initial velocity along the Z-axis,the motion will be in the `X-Y` plane.Also, its speed remains unchanged .As the magnetic field is not uniform, the particle does not go along a circle.
The force at time `t` is `vecF=qvecvxxvecB`
`=q(veciv_(x)+vecjv_(y))xx((mu_(0)i)/(2pix)K)`
`=-vecjqv_(x)(mu_(0)i)/(2pix)+veciqv_(y)(mu_(0)i)/(2pix)`
Thus,`a_(x)=F_(x)/m=(mu_(0)qi)/(2pim)(v_(y))/x=lambdav_(y)/x`..(i)
where `lambda=(mu_(0)qi)/(2pim)`
Also, `a_(x)(dv_(x))/(dt)=(dv_(x))/(dt)(dx)/(dt)=(v_(x)dv_(x))/(dx)`...(ii)
As, `v_(x)^(2)+v_(y)^(2)=v^(2)`...(iii)
From (i),(ii) and (iii)
`(d_(y)dv_(y))/(dx)=(-lambdav_(y))/x`
or `(dx)/x=(-dv_(y))/lambda`
initially `x=x_(0)` and `v_(y)=0`.At minimum separation from the wire, `v_(x)=0` so tht `v_(y)=v`.
Thus `underset("x0")overset("x")(oint)(dx)/x=underset(0)overset(v)(oint)(dv_(y))/-lambda`
or "in" `x/x_(0)=-v/lambda`
or `x=x_(0)e^(v//lambda)=x_(0)e^-((2pimv)/(mu_(0)qi))`