From the surface of a round wire of radius a carrying a direct current `I` an electron escapes with a velocity `v_(0)` perpendicular to the surface. Find what will be the maximum distance of the electron from the axis of the wire before it turns back due to the action of the magnetic field generated by the current.

Explanation

Choose the wire along the z-axis, and the initial direction of the electron, along the x-axis. Then the magnetic field in the `x-z` plane is along the y-axis and outside the wire it is,
`B = B_(y) = (mu_(0) I)/(2pi x), (B_(x) = B_(x) = 0`, if `y = 0`)
The motion must be condined to the `x-z` plane. Then teh equacations of motion are,
`(d)/(dt) mv_(x) = -ev_(z) B_(y)`
`(d(mv_(x)))/(dt) = + ev_(x) B_(y)`
Multiplying the first equacation by `v_(x)` and then adding,
`v_(x) (dv_(x))/(dt) + v_(x) (dv_(z))/(dt) = 0`
or, `v_(x)^(2) + v_(z)^(2) = v_(0)^(2)`, say, or, `v_(z) = sqrt(v_(0)^(2) - v_(x)^(2))`
Then, `v_(x) (dv_(x))/(dx) = - (e).(m) sqrt(v_(0)^(2) - v_(x)^(2)) (mu_(0) I)/(2pi x)`
or, `- (v_(x) dv_(x))/(sqrt(v_(0)^(2) - v_(x)^(2))) = (mu_(0) I e)/(2pi m) (dx)/(x)`
Intergating `sqrt(v_(0)^(2) - v_(x)^(2)) = (mu_(0) Ie)/(2pi m)` In `(x)/(a)`
on using `v_(x) = v_(0)`, If `x = a` (i.e. initially).
Now, `v_(x) = 0`, when `x = x_(m)`,
so, `x_(m) = a e^(v_(0)//b)`, where `b = (mu_(0) Ie)/(2pi m)`