A particle of mass m and charge q is moving in a region where uniform, constant electric and mangetic fields `vec E and vec B` are present. `vec E and vec B` are parallel to each other. At time `t=0,` the velocity `vec v_0` of the particle is perpendicular to `vec E` (Assume that its speed is always `lt lt c`, the speed of light in vacuum). Find the velocity `vec v` of the particle at time `t`. You must express your answer in terms of `t, q, m,` the vector `vec v_0, vec E` and `vec B` and their magnitudes `vec v_0, vec E` and `vec B`.

Answer

`vec=vecv_(0)cos ((qBt)/m)+(qtvecE)/m+[((vecv_(0)xxvecB))/B sin ((qBt)/m)]`

Explanation

`vecv=vecv_(0)cos ((qBt)/m)+(qtvecE)/m+[((vecv_(0)xxvecB))/Bsin ((qBt)/m)]`
Let velocity of the particle be `v_(0)` in x direction and B is in y-direction and `vecv` at time t
`vecF=q(vecE+vecvxvecB),vecv=v_(x)hati+v_(y)hatj+v_(z)hatk,vecF=m((dv_(x))/(dt)hati+(dv_(y))/(dt)hatj+(dv_(z))/(dt)hatk)`
`=q(Ehatj+(v_(x)hati+v_(y)hatj+v_(z)hatk)xxvecBhatj), m(dv_(x))/(dt)=qv_(z)B,m(dv_(y))/(dt)=dE,m(dv_(z))/(dt)=qv_(x)B`
`m(d^(2)v_(x))/(dt^(2))=-qB(dv_(z))/(dt),m(d^(2)v_(x))/(dt^(2))=-qB(dBv_(x))/m`
`(d^(2)v_(x))/(dt^(2))+((qB)/m)^(2)v_(x)=0`

`v_(z)=v_(y)=0, v_(x)=v_(0)` at time `t=0`
`v_(x)=v_(0)"cos"(qB)/mt:v_(y)=((qE)/m)t`
`v_(z)=v_(0)sin ((qB)/m)t`
Velocity of the particle `vecv=vecv_(0)cos ((qBt)/m)+(qtvecE)/m+[((vecv_(0)xxvecB))/Bsin ((qBt)/m)]`