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$$eE=evB$$ $$\Rightarrow v=\frac{E}{B}$$
Note that maximum magnetic force applies as velocity and magnetic field vectors are perpendicular to each other. Substituting expression of v in the kinetic energy equation obtained earlier, we have :
$$\frac{m{E}^{2}}{{B}^{2}}=2eV$$ $$\Rightarrow \alpha =\frac{e}{m}=\frac{{E}^{2}}{2V{B}^{2}}$$
All the quantities on the right hand side of the equation are measurable, allowing us to measure the specific charge of electron. As a matter of fact, the determination of specific charge of particles composing cathode ray by J.J.Thomson is considered to be the discovery of electron. It can also be easily inferred that he could determine the nature of charge of an electron by studying direction of deviation (upward or downward) when only either of the fields operate. In the derivation above, we measure potential difference applied to accelerate particle between cathode and anode. We should, however, realize that we can determine specific charge measuring some other quantities as well. We can measure the deflection of electron beam when either of two fields operates and use the data to determine specific charge of an electron.
Problem : The d.c. voltage applied to accelerate particle between cathode and anode and the d.c. voltage applied to the plates to produce electric field perpendicular to electrons beam are equal in the Thomson’s experimental set up. If each of the two d.c. voltages as applied are doubled, then by what factor should the magnetic field be changed to keep the electron beam un-deflected.
Solution : Let ${V}_{1}$ , ${E}_{1}$ and ${B}_{1}$ be the potential difference, electric field and magnetic field for un-deflected condition. Then, the specific charge is given by :
$$\alpha =\frac{e}{m}=\frac{{E}_{1}^{2}}{2{V}_{1}{B}_{1}^{2}}$$
Here, the electric field can be expressed in terms of potential difference provided we know the separation between plates. Let the separation be d.
$${E}_{1}=\frac{{V}_{1}}{d}$$
Putting in the equation above, we have :
$$\Rightarrow \alpha =\frac{{V}_{1}^{2}}{2{d}^{2}{V}_{1}{B}_{1}^{2}}=\frac{{V}_{1}}{2{d}^{2}{B}_{1}^{2}}$$
Let ${B}_{2}$ be the new magnetic field when two potential differences as applied are doubled. Here,
$${V}_{2}=2{V}_{1}$$
Putting new values in the expression for specific charge (note that specific charge of electron is a constant),
$$\alpha =\frac{2{V}_{1}}{2{d}^{2}{B}_{2}^{2}}$$
Combining two equations,
$$\frac{2{V}_{1}}{2{d}^{2}{B}_{2}^{2}}=\frac{{V}_{1}}{2{d}^{2}{B}_{1}^{2}}$$ $$\Rightarrow 2{V}_{1}{d}^{2}{B}_{2}^{2}=4{V}_{1}{d}^{2}{B}_{1}^{2}$$ $$\Rightarrow \frac{{B}_{2}^{2}}{{B}_{1}^{2}}=2$$ $$\Rightarrow \frac{{B}_{2}}{{B}_{1}}=\sqrt{2}$$
Once the magnetic and electric forces are balanced, electric field is switched off and electron beam is allowed to be deviated due to magnetic field. The magnetic force acts always perpendicular to the direction of motion. The particle, therefore, moves along a circular path inside the region of magnetic field. When electron moves out of the magnetic field, it moves along the straight line and hits the fluorescent screen. If R be the radius of curvature, then :
$$\frac{m{v}^{2}}{R}=evB$$ $$\Rightarrow mv=eRB$$
Substituting $v=E/B$ as obtained earlier
$$\alpha =\frac{e}{m}=\frac{E}{R{B}^{2}}$$
We measure R using geometry. We see that the angles enclosed between pairs of two perpendicular lines are equal. Hence,
$$\phi =\frac{DG}{R}=\frac{OI}{FO}$$ $$\Rightarrow R=\frac{FOXDG}{OI}$$
We approximate DG to be equal to the width of magnetic region.
In this case, once the magnetic and electric forces are balanced, electric field is switched off and electron beam is allowed to be deviated due to electric field. The electron beam moving into the region of electric field experiences an upward force. The force in upward (y-direction) imparts acceleration in y-direction. The particle, however, moves with same velocity in x-direction. As a result, path of motion is parabolic. Let the length of plate be L and y be the deflection inside the plate. Then, time to travel through the plate is :
$$t=\frac{L}{v}$$
and acceleration of the particle in y-direction is :
$${a}_{y}=\frac{{F}_{E}}{m}=\frac{eE}{m}$$
The vertical displacement is :
$$y=\frac{1}{2}{a}_{y}{t}^{2}$$
Substituting for time and acceleration, we have :
$$y=\frac{1}{2}X\frac{eE}{m}X\frac{{L}^{2}}{{v}^{2}}=\frac{eE{L}^{2}}{2m{v}^{2}}$$
Substituting v=E/B as obtained earlier,
$$y=\frac{eE{L}^{2}}{2m}X{\left(\frac{B}{E}\right)}^{2}=\frac{\alpha E{L}^{2}}{2}X{\left(\frac{B}{E}\right)}^{2}$$ $$\Rightarrow \alpha =\frac{e}{m}=\frac{2yE}{{B}^{2}{L}^{2}}$$
It is clear that measuring GH and HI, we can determine angle φ and then y as required.
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