The region between ` x=o and x = L` is filled with uniform, steady magnetic field `B_(0) hat(k)`. A particle of mass `m` , positive charge `q` and velocity `v_(0) hat(i)` travels along ` x-axis` and enters the region of the magnetic field. Neglect gravity throughout the question .
(a) Find the value of `L` if the particle emerges from the region of magnetic field with its final velocity at angle `30^(@)` to its initial velocity.
(b) Find the velocity of the particle and the time spent by it in the magnetic field, if the magnetic field now extends upto `2.1L`.

Answer

(a)`(mv_(0))/(2qB)` (b)`-v_(0)hati(pim)/(qB)`

Explanation

(a)`theta=30^(@)`
`sintheta=L/R`
Here,`R=(mV_(0))/(B_(0)q)`
`therefore sin30^(@)=L/((mV_(0))/(B_(0)q))`
or `1/2=(B_(0)q)/(mV_(0))`
`therefore L=(mV_(0))/(2B_(0)q)`
(b)In part (a)
`sin30^(@)=L/R`
or `1/2=L/R`
or `L=R//2`
Now when `L=2.1 L` or `2.1/2 R`
`rArr L gt R`
Therefore, deviation of the particle is `theta=180^(@)` is as shown.
`therefore vecV_(f)=-V_(0)hati-vecV_(B)`
and `t_(AB)=T//2=(pim)/(B_(0)q)`