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 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(v_(x)veci+v_(y)hatj)xx(-(mu_(0)i)/(2pix)veck)`
`=hatjqv_(x)=(mu_(0)i)/(2pix)-veciqv_(y)(mu_(0)i)/(2pix)`
Thus `a_(x)=(F_(x))/(m)=-(mu_(0)qi)/(2pim)(v_(y))/(x)=-lambda(v_(y))/(x)...(i)`
where `lambda=(mu_(0)qi)/(2pim)`
Also `a_(x)=(dv_(x))/(dt)=(dv_(x))/(dx)(dx)/(dt)=(v_(x)dv_(x))/(dx)...(ii)`
As, `v_(x)^(2)+v_(y)^(2)=v^(2)`,
giving `v_(x)dv_(x)=-v_(y)dv_(y)...(iii)`
From (i), (ii) and (iii),
`(v_(y)dv_(y))/(dx)=(lambdav_(y))/(x)`
or, `(dx)/(x)=(dv_(y))/(lambda)`
Initially `x=x_(0)andv_(y)=0`. At minimum separation from the wire, `v_(x)=0` so that `v_(y)=-v`.
Thus `underset(x_(0))overset(x)int(dx)/(x)=underset(0)overset(-v)int(dv_(y))/(lambda)or,In(x)/(x_(0))=-(v)/(lambda)`
or, `x=x_(0)e^(-v//lambda)=x_(0)e^(-(2pimv)/(mu_(0)qi))`