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

Solution: Let the particle be initially at P. 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
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 `( vec F)= (vec qv) xx (vec B)`
` = q((vec i)vx + (vec j)vy) xx (- (mu_0)i/2 pi x) vec k)`
` = ((vec j) q(v_x)) ((mu_0)i/2pi x) - (( vec i ) (qv_y)((mu_0)i/2 pi x)`
Thus a_x= (F_x/m)= -((mu_0)qi/2 pi m) ((v_y)/x) = -lambda(V_y/x) .......(i)`
` where lambda = ((mu_0)qi/2pi m).`
` 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,`
` (2v_x)(dv_x) + ((2v_y) (dv_y))=0`
` giving ((v_x)(dv_x))= - (v_y) (dv_y).....(iii) `
` From (i), (ii) and (iii), `
` ((v_y)(dv_y)/dx)= (lambda(v_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 that v_y = -v .`
` Thus (int_(x_(0))^(x))(dx/x)= ((int_(0))^(-v)) ((dv_y)/lambda)`
` or, In (x/x_0)=(-v/lambda)`
` or, x = (x_0) (e^(-v/lambda))= (x_0) (e ^(- 2pi m v /((mu_0)qi))