A particle of mass m and charge q is moving in a region where uniform, constant electric and magnetic field `vecE` are `vecB` parallel to each other. At time t = 0 the velocity `vecv_(0)` of the particle is perpendicular to `vecE`. (Assume that its speed is always << c, the speed of light in vacuum). Find the velocity `vecv` of the particle at time t. You must answer in terms of t, q, m the velocity `vecv_(0),vecE` and `vecB` . , and their magnitude `v_(0),E` and B. ___________.

Answer

`vecv_((t))=cos omegat vecv_(0)+((sinomegat)(vec v_(0)xxvecB_(0)))/B+("gt")/mvecE` where `omega=qB//m`

Explanation

`vecv-v_(x)hati+v_(y)hatj+v_(2)hatk`
`vecv_(x)=(qE)/mt,v_(y)=v_(0) cos ((gB)/mt)`
`v_(z)=-v_(0) sin ((gB)/mt)`
`i=vecE/|vecE|=vecB/|vecB|`
`hatj=vecv_(0)/|vecv_(0)|, hatk=(-vecv_(0)xxvecB)/(|vecv_(0)xxvecB|)=(vecvxxvecE)/(|vecv_(0)xxvecE|)`
(ii)`hatj=vecE/E or vecB/B,hati=vecv_(0)/v_(0)`
`hatk=(vecv_(0)vecB)/(v_(0)B)`
Force due to electric field will be along `y`-axis. Magnetic force will not affect the motion of charged particle in the direction of electric field (or `y-`axis )So
`a_(y)=(F_(e))/m=(qE)/m="constant therefore",v_(y)=a_(y) t=(qE)/m t`..(1)
The charged particle under the action of magnetic field describes a circle in `x-y` plane (perpendicular to `vecB`) with
`T=(2pim)/(Bq)` or `omega=(2pi)/T=(qE)/m`
Initially `(t=0)` velocity was along `x`-axis.Therefore, magnetic force `(vecFm)` will be along positive `z`-axis `[vecF m=q(vecv_(0)xxvecB)]`.Let it makes an angle `theta` with `x`-axis at time `i`. then `theta=omegat`
`thereforev_(x)=v_(0) cos omegat=v_(0) cos ((qB)/mt)`and ..(2)
`v_(z)=v_(0) sin omegat=v_(0)sin((qB)/mt)`..(3)
From (1),(2) and (3)
`therefore vecv=v_(x)+v_(y)hatj+v_(z)hatk`
`therefore vecv=v_(0)cos ((gB)/mt)(vecv_(0)/v_(0))+(qB)/mt(vecE/E)+v_(0)sin((gB)/mt)((vecv_(0)xxvecB)/(v_(0)B))`
or `vecv=cos((gB)/mt)(vecv_(0))+(g/mt)(vecE)+sin ((gB)/mt) ((vecv_(0)xxvecB)/(v_(0)B))`
`rArr` The path of particle will be a helix of incresing pitch.The axis of the helix will be along `y` axis.