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An electron travels a distance of 0.10 m in anelectric field of intensity 3200 V/m, entersperpendicular to the field with a velocity 4  10^7m/s,what is its deviation in its path :(1) 1.76 mm. (2) 17.6 mm.(3) 176 mm. (4) 0.176 mm.

Aadhi , 8 Years ago
Grade 12th pass
anser 1 Answers
Vikas TU

Last Activity: 8 Years ago

In y direcn.
y = 0.5*(eE/m)*t^2
put the value of charge mass and given electric field inensity
and get the reln. in y and t.
 
Calculate t from:
x = v*t
0.10 = 4*10^7*t
 
and put it in y eqn. and the deviation value .i.e. y.

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