From the surface of a round wire of radius a carrying a direct current i, a positive charge q escapes with a velocity `v_0` perpendicular to the surface. Find what the maximum distance of the electron will ne from the axis of the wire before it turns back due to the action of the magnetic field generated by the current.

Explanation

Let at any instant velocity of charge is v, then force on the charge due to magnetic field caused by wire is given by
`vecF=q(vecvxxvecB)`
`ma_xhati+ma_yhatj=q(v_x hati+v_zhatk)xx(mu_0I)/(2pix)hatj`
`ma_xhati+ma_yhatj=qv_x(mu_0I)/(2pix)hatk-qv_z(mu_0I)/(2pix)hati`

`implies ma_x=-(qmu_0I)/(2pi) (v_z)/x`
`implies a_x=-(qmu_0I)/(2pim) (v_z)/x`
["But" `v_0^2=v_x^2impliesv_z=+-sqrt(v_0^2-v_x^2)`
Here we have to take `v_z=+sqrt(v_0^2-v_x^2`) because `a_x` is negative]
`implies v_x(dv_x)/(dx)=-(qmu_0I)/(2pim) (sqrt(v_0^2-v_x^2))/x`
`implies int_(v_0)^0(v_xdv_x)/(sqrt(v_0^2-v_x^2))=(qmu_0I)/(2pim)int_a^(x_(max)) (dx)/x`
`implies [-sqrt(v_0^2-v_x^2)]_(v_0)^0=-(qmu_0I)/(2pim) ln((x_(max)/a)`
`implies v_0=(qmu_0I)/(2pim) ln((x_(max)/a)`
`implies x_(max)=ae^((2pimv_0)/(qmu_0I))`
when x is miximum, x component of velocity is zero.