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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 line x=0 and x=a. Beyond the line x=a (i.e., in the region x>a), there exists a magnetic field of strength B, directed along the positive y-axis. Find
at which point does the particle meet the line x=a
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Solution
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The work done by the potential difference gets stored as its kinetic energy.
12mv2=qVv=(2qVm)1/2 (i)
When it enters the region xϵ[0,a), it experiences an electric field E=E^j
Time taken to cross the region: vt=a
t=av=a(m2qV)1/2 (ii)
The distance travelled in y-direction during this time is
y=12qEmt2=12×qEm×a2×m2qV
y=14Ea2V
Hence, the particle meets the line x=a at point
(x,y)=(a,14Ea2V)

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