A potential difference of 600 V is applied across the plates of a parallel plate capacitor. 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. What is the magnitude of the magnetic field between the capacitor plates?
Electric intensity E=Vd
where V is the potential difference between the plates and d, the separation between them.
d = 3 mm = 3×10−3m
E=Vd=6003×10−3=2×105 V m−1
Since the electron moves undeflected between the plates, the force clue to magnetic field must balance the force due to electric field.
B e ν=e E or B=Eν=2×1052×106=0.1 T