. A potential difference of 60 volts 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 x 10 6 m/sec moves undeflected between the plates. Find the magnitude and direction of the magnetic field in the region between the condenser plates. (Neglect the edge effects). (Charge of the electron) = - 1.6 x 10 -19 coulomb

Question

Navjyot Kalra · Answer

Hello Student,

Please find the answer to your question

The force on electron will be towards the left plate due to electric field and will be equal to

F_e = eE

NOTE : For the electron to move undeflected between the plates there should be a force (magnetic) which is equal to the electric force and opposite in direction. The force should be directed towards the right as the electric force is towards the left.

On applying Fleming’s left hand rule we find the magnetic field should be directed perpendicular to the plane of paper inwards. Therefore,

Force due to electric field = Force due to magnetic field.

eE = evB

∴ B = E / v = V /d / v [∵ E = V /d]

Where V = p.d. between plates

d = distance between plates

∴ B = 600 / 3 x 10^-3 / 2 x 10⁶ = 600 / 3 x 10^-19 x 2 x 10⁶

B = 0.1 Tesla

Thanks
Navjot Kalra
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