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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 2×106ms−1 moves undeflected between the plates. What is the magnitude of the magnetic field between the capacitor plates?

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Detailed Solution

Electric field E=Vdwhere V is the potential difference between the plates and d, the separation between them.d=3mm=3×10−3mE=Vd=6003×10−3=2×105Vm−1Since the electron moves undeflected between the plates, the force due to magnetic field must balance the force due to electric field. ThusBev=eEB=Ev=2×1052×106=0.1T

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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 2×106ms−1 moves undeflected between the plates. What is the magnitude of the magnetic field between the capacitor plates?