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A positively charged particle, having ch...

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.)

Text Solution

Verified by Experts

The correct Answer is:
B

a. Speed of particle at origin `v=sqrt((2qV)/m)=v_x`
`t=x/v_x=a/sqrt((2qV)/m)=sqrt(m/(2qV))`
`y=1/2a_yt^2=1/2[(qE)/m][a^2xxm/(2qV0]=(a^2E)/(4V)`
b. Component parallel to `B` is
`v_y=a_yt=(qE)/(m)(asqrt(m/(2qV)))=Easqrt(1/(2mV))`
Now, pitch= component parallel to `Bxx time` period.
`=v_yT=(Easqrt(q/(2mV))((2pim)/(Bq))`
`=(piEa)/B sqrt((2m)/(qV))`
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