Question

In Fig. 28-58, an electron of mass m, charge -e, and low (negligible) speed enters the region between two plates of potential difference V and plate separation d, initially headed directly toward the top plate. A uniform magnetic field of magnitude B is normal to the plane of the figure. Find the minimum value of B, such that the electron will not strike the top plate.

Short answer

The minimum value of B such that the electron will not strike the top plate is
$B=\sqrt{\frac{mV}{2e{d}^2}}$

Step-by-step solution

Given

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.

Determine the concept and the formulas

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. $PE=qV$
   

Calculate the minimum value of B such that the electron will not strike the top plate.

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:

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