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(Fig. 29-16).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 negative z axis. The magnetic force on the particle is, therefore, in the xy plane. As there is no initial velocity along the z axis, the motion will be in the xy plane. We know that the magnetic field does not do any work. So the speed of the charged particle does not change as it moves through the magnetic field. Since the magnetic field is not uniform, the particle does not go along a circle but rather a curved path in the xy plane.
Calculations : The force at time t is `vecF=qvecvxxvecB`. Here we use the basic definition of force and velocity.
`vecF=q(v_(x)hati+v_(y)hatj)xx(-(mu_(0)i)/(2pix)hatk)`
`=qv_(x)(mu_(0)i)/(2pix)hatj-qy_(y)(mu_(0)i)/(2pix)hati`.
We can resolve this force along x and y directions. Thus
`a_(x)=(F_(x))/(m)=(mu_(0)qi)/(2pim)(v_(y))/(x)=-lambda(v_(y))/(x)`,
where
`lambda=(mu_(0)qi)/(2pim)`
Also, `a_(x)=(dv_(x))/(dt)=(dv_(x))/(dx)(dx)/(dt)=(v_(x)dv_(x))/(dx)`

Since the speed remains constant,
`v_(x)^(2)+v_(y)^(2)=v^(2)`
Differentiating this expression, we get
`v_(x)dv_(x)+2v_(y)dv_(y)=0`
`v_(x)dv_(x)=-v_(y)dv_(y)`
From Eqs. 29-26 and 29-27, we have
`(v_(y)dv_(y))/(dx)=(lambdav_(y))/(x)`, `(dx)/(x)=(dv_(y))/(lambda)`
Initially `x=x_(0)` and `v_(y)=0` . When x is minimum, by the concept of minima, `v_(x)=0` so that `v_(y)=-v`. Thus
`int_(x_(0))^(x)(dx)/(x)=int_(0)^(-v)(dv_(y))/(lambda)`
Solving, we get
`ln .(x)/(x_(0))=-(v)/(lambda)`
`x=x_(0)e^(-v//lambda)=x_(0)e^(-2pimv//mu_(0)qi)`
The charged particle will turn back without changing its speed. Even if a charged particle, say a stray electrons , is emitted from the wire, it will move in a curved path and turn back at some point. However, in that case also, the path of the particle will not be circular.