A particle of mass m and charge + q is projected from origin with velocity `vec(V)=V_(0)hati` in a magnetic field `vec(B)=-(B_(0)x)hatk.` Here `V_(0)` and `B_(0)` are positive constants of proper dimensions. Find the radius of curvature of the path of the particle when it reaches maximum positive x co-ordinate.

Answer

2

Explanation

Magnetic force on the particle at origin is along positive `y-` direction, so its trajectory will be concave upwards. As the magnetic field is nonuniform, trajectory is not circular. Magnetic force is always perpendicular to velocity . So its magnitude remains constant. Let at point `P(x,y)` its velocity vector make an angle `theta` with positive `x-` axis . Then the magnetic force `vec(F_(B))` will be at angle`theta`with the positive `y-` direction . So,
`a_(y)=((F_(B))/(m))cos theta(or)(d upsilon_(y))/(dt)=((B_(0)x)(q upsilon_(0)cos theta))/(m)`
`[F_(B)=Bq upsilon_(0)sin90^(@)]` or
`((d upsilon_(y))/(dx)).((dx)/(dt))=((B_(0)qx)/m)(upsilon_(0)cos theta) ` where
`(dx)/(dt)=upsilon_(x)=upsilon_(0)cos theta`
Thus`(d upsilon_(y))/(dx)=((B_(0)q)/(m))x`
At maximum `x-` displacement , velocity is along positive `y-` direction
or `" "int_(0)^(upsilon_(0))dupsilon_(y)=((B_(0)q)/(m))int_(0)^(x_(max))x dx`
or`" "upsilon_(0)=((B_(0)q)/(m))((x_(max)^(2))/(2))`
or `" "x_(max)=sqrt((2m upsilon_(0))/(B_(0)q))`