A potential difference of 600 V is applied across the plates of a parallel plate condenser. The separation between the plates is 3 mm. An electron projected vertically, parallel to the plates, with a velocity of 2×106 ms-1 moves undeflected between the plates. Find the magnitude of the magnetic field in the region between the condenser plates. (Neglect the edge effects. Charge of the electron= -1.6×10−19C )
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a
10 T
b
0.1 T
c
0.01 T
d
1 T
answer is B.
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Detailed Solution
The force on electron will be toward the left plane due to electric field and will be equal to Fe=eE.For no deflection, electric force should be balanced by magnetic force.eE=evB⇒B=Ev=V/dvWhere V =potential difference between the plates, andd=distance between the platesB=600/3×10−32×106=6003×10−3×2×106 ⇒B=0.1 T .