A non-uniform magnetic field `vecB=B_(0)(1+y/d)(-hatk)` is present in region of space in between `y=0 & y=d`.The lines are shown in the diagram.A particle of mass `m` and positive charge `q` is moving.Given an initial velocity `vecv=v_(0)hati`.Find the components of velocity of the particle when it leaves the field.

Answer

`V_(x)=V_(0)-(3qB_(0)d)/(2m)` and `V_(y)=sqrt(V_(0)^(2)-(V_(0)-(3qB_(0))/(2m)d))^(2)`

Explanation

Let at time `t` particle be at point `P(x,y)` and its velocity be
`vecB=(V_(x)hati+V_(y)hatj)`
`|vecV|=|vecV_(0)|rArr V_(0)^(2)=V_(x)^(2)+V_(y)^(2)`
(work done by magnetic field is always zero so charge in magnitude of velocity)Then, magnetic force on the particle at point `P` is
`vecF=q(V_(x)hati+V_(y)hatj).B_(0)(1+y/d)(-hatk) rArr -qB_(0)[1+y/d]dy=mdv_(x)`
Now when the particle will be coming out of the at that point `y=d`.Let the velocity in `x`-direction be `V_(x)` then integrating we get,
`underset(V_(0))overset(V_(x))intdv_(x)=-(qB_(0))/m underset(0)overset(d)intdy=-(qB_(0))/m[d+d^(2)/(2d)]=-(3qB_(0)d)/(2m)`
so `V_(x)=V_(0)-(3qB_(0)d)/(2m)`
`rArr V_(y)=sqrt(V_(0)^(2)-V_(x)^(2))`
`rArrV_(y)=sqrt(V_(0)^(2)-(V_(0)-(3qB_(0))/(2m)d))^(2)`