A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

Explanation

(a) Since given charged particle passes undeflected through a given unidirectional magnetic field, magnetic force exerted on it must be zero.
`thereforeF_(m)=0`
`thereforeqvBsintheta=0`
`thereforesintheta=0" "(becauseqne0,vne0,Bne0)`
`thereforetheta=0^(@)ortheta=180^(@)`
`vecv` would be in the direction of `vecB` or opposite to direction of `vecB`.
(b) Yes, here final speed of charged particle would be same as initial speed. Because initial velocity `vecv` can be decomposed into two mutually perpendicular components as follows:

(i) Here `v_(||)` is in the direction of `vecB` and so `vecv_(||)xxvecB=0`
Magnetic force dur to `v_(||)=0`
No work will be done by magnetic force.
`(becauseF_(m)=0)`
(ii) Here `v_(_|_)` is perpendicular to `vecB`. Hence, displacement of charged particle will be perpendicular to magnetic force. (`because` Direction of `vecF_(m)` in this case is radial and direction of displacement is tangential) and so no work is done by magnetic force.
Thus, when any charged particle moves in any magnetic field in any direction, no work is done on it. Hence, `W=DeltaK=0rArrK="constant"rArr1/2mv^(2)="constant"rArrv="constant"`
( c)
When an electron given in the statement passes undeflected through a region containing electric as well as magnetic field, Lorentz force on it must be zero. Hence,
`vecF=vecF_(e)+vecF_(m)`
`thereforevec0=qvecE+q(vecvxxvecB)`
`thereforevec0=(-e)vecE+(-e)(vecvxxvecB)`
`thereforeevecE=e(vecBxxvecv)`
`thereforevecE=vecBxxvecv" "("For electron")" "...(1)`
From equation (1), direction of `vecB` should be such that direction of `vecBxxvecv` for electron remains in the direction of `vecE`. That is why magnetic field `vecB` is shown in the diagram in vertically downward so that angle `theta` between `vecvandvecB` would be `90^(@)`.
From equation (1),
`E=Bvsintheta`
`thereforeE=Bvsin90^(@)`
`thereforeE=Bv`
`thereforeB=E/v" "...(2)`
Thus, for the condition made in the statement, magnetic field should be applied in vertically downward direction with magnitude given by equation (2).