In Figure below, an electron accelerated from rest through potential difference

Question

In Figure below, an electron accelerated from rest through potential difference V1=0.920 kV enters the gap between two parallel plates having separation d = 21.2 mm and potential difference V2= 90.8 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. In unit-vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap?

I got....

(0i+0j+_____k)mT

I need what goes in front of k

Explanation / Answer

The potential difference between the plates will exert a force on the electron; As the lower plate is at a lower potential the force will be upwards towards y; This force will be equal to (V2*e)/d; d = 21.2 mm = 0.0212 m

Hence we need a magnetic force to counter this force
Magnetic force = qvB where q is charge; v is velocity and B is field strength
In this case Kinetic energy after it has been accelerated = 920*e
Hence (1/2)mv2 = 920*e
Hence v = (1840*e/m)

Hence magnetic force = e*(1840*e/m)*B
Hence to balance and make electron move undeflected we have;
e*(1840*e/m)*B = (V2*e)/d
Hence B = V2/(d* (1840*e/m)) = 90.8/(0.0212 * (1840*(1.7588* 1011) = 23.808 * 10-5 T = 0.23808 mT

Now by Flemming's left hand rule, we know that for the magnetic force to oppose the electric force (i.e. to be downward) ; the magnetic feild should exist in the positive z direction;

Hence co efficient of k will be 0.23808

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