A positively charged particle, having charge q, is accelerated by a potential difference V. This particle moving along the x-axis enters a region where an electric field E exists. The direction of the electric field is along positive y-axis. The electric field exists in the region bounded by the lines `x=0 and x=a.` Beyond the line `x=a` (i.e., in the region `xgta`), there exists a magnetic field of strength B, directed along the positive y-axis. Find

a. at which point does the particle meet the line `x=a`.
b. the pitch of the helix formed after the particle enters the region `xgea.` (Mass of the particle is m.)

Answer

(a) `(x,y)=(a,1/4 (Ea^2)/V)`; (b) `p=(piaE)/B((2m)/(Vq))^(1//2)`

Explanation

The work done by the potential difference gets stored as its kinetic
energy.
`:. 1/2mv^2 = qV implies v = ((2qV)/m)^(1//2)....(i)`
(a) When it enters the region `xepsilon[0,a)`, it experiences an electric
field `vecE=Ehatj`
Time taken to cross the region:`vt=a`
`implies t=a/v=a(m/(2qV))^(1//2)....(ii)`
The distance travelled in y-direction during this time is
`y=1/2(qE)/mt^2=1/2xx(qE)/mxxa^2xxm/(2qV)`
`implies y=1/4(Ea^2)/V`
Hence, the particle meets the line `x=a` at point
`(x,y)=(a,1/4(Ea^2)/V)`
(b) Now, velocity of the particle as it crosess the line `x=a`
`vecv=((2qV)/m)^(1//2)hati+(qE)/m((2qV)/m)^(1//2)hatj`
Magnetic field in this region, `vecB=Bvecj`
Hence, the time period of revolution ,`t=(2pim)/(qB)`
Pitch `P=v_(||)t=(2pim)/(mqB)xxaE((mq)/(2V))^(1//2)`
`P= (piaE)/B((2m)/(Vq))^(1//2)`.