A particle having mass m and charge q is released from the origin in a region in which electric field and magnetic field are given by
` vecB =- B_0vecJ and vecE = E_0 vecK.`
Find the speed of the particle as a function of its z-coordinate.

Answer

`y=(E_(0)m)/(qB_(0)^(2))[1-cos (qB_(0))/mt]`

Explanation

`m(dv)/(dt)=qE_(0) hatj[v_(x)hati+v_(y)hatj]B_(0)hatk`
`m(dv)_(y)/(dt)hatj+m(dv)_(x)/(dt)hati=[qE_(0)-qv_(x)B_(0)] hatj+qv_(y)B_(0)hati`
`m(dv)_(y)/(dt)=[qE_(0)-qv_(x)B_(0)] -(1)`
`m(dv)_(y)/(dt)=qv_(x)B_(0)` --(2)
From (1)`V_(x)=[qE_(0)-m"dv"_(y)/(dt)]1/(qE_(0))`
From (2) `m/(qE_(0)) d/(dt)[qE_(0)-m"dv"_(y)/(dt)]=qv_(y)B_(0)`
`-(d^(2)v_(y))/"dt"^(2)=(q^(2)v_(y)B_(0)^(2))/m^(2) or (d^(2)v_(y))/"dt"^(2)+(q^(2)v_(y)B_(0)^(2))/m^(2)=0`
Solution of above equation:
`v_(y)=A sin (omega t +phi)`-(3)
where `omega=(q B_(0))/m` at `t=0,v_(y)=0,phi=0 , v_(y)=A sin omega t`
at `t=0,a=(q B_(0))/m a="dv"_(y)/(dt)=A omega cos omega t (qE_(0))/m=Axx(qB_(0))/m rArr A =E_(0)/B_(0)`
This equation (3) `v_(y)=E_(0)/B_(0)sin omega t(dy)/(dt) =E_(0)/B_(0)sin omega t rArr y=[-E_(0)/B_(0)cos omega t]_(0)^(t)`
`y=(E_(0)m)/(B_(0)xxq B_(0))[1-cos omegat] rArr y=(E_(0)m)/(B_(0)^(2))[1-"cos" (q B_(0))/mt]`