91Ó°ÊÓ

1. The mass of electron is m.
2. The charge on electron is -e.
3. Potential difference across the plates is V.
4. The distance of plate separation is d.
5. The magnitude of magnetic field is B.

Consider the equations of electric force and magnetic force. The magnetic force opposes the motion of electron, so to prevent the electron from striking, the magnetic force must be greater than the electric force.

Formulae:

1. $F_B=qvB$
   
2. $F_E=\frac{qV}{d}$
   
3. $KE=\frac{1}{2}m{v}^2$
   
4. $PE=qV$
   

Consider the equation for potential energy of electron just before it strikes the top plate as,

$U=qE=\frac{eV}{d}$

Now, according to the conservation of energy principle is as follows:

KE=PE

So,

$\frac{1}{2}m{v}^2=eV$

Rearranging for velocity derive the equation as:

$v=\sqrt{\frac{2eV}{m}}$

Now, consider the equation for magnetic force as

$F_B=qvB=evB$

Substituting for the velocity, we get

$F_B=eB\sqrt{\frac{2eV}{m}}$

Now, to prevent the electron from striking the top plate, F_B must be greater than F_E.

So, to find the minimum magnetic field, let us assume those to be equal.

Hence,

F_B=F_E

$eB\sqrt{\frac{2eV}{m}}=\frac{eV}{d}$

Rearranging it for magnetic field, solve as:

$\begin{array}{l}B=\frac{V}{d}×\sqrt{\frac{m}{2eV}}\\ B=\sqrt{\frac{mV}{2e{d}^2}}\end{array}$