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Magnetic Field — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsMagnetic Field5 Marks
Question
Electrons emitted with negligible speed from an electron gun are accelerated through a potential difference V along the x-axis. These electrons emerge from a narrow hole into a uniform magnetic field B directed along this axis. However, some of the electrons emerging from the hole make slightly divergent angles, as shown in the figure. Show that these paraxial electrons are refocussed on the x-axis at a distance $\sqrt{\frac{8\pi^2\text{mV}}{\text{eB}^2}}.$
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Answer

Given magnetic field = B, Pd = V, mass of electron = m, Charge = q,
Let electric field be ‘E’
$\therefore\text{E}=\frac{\text{V}}{\text{R}'}$
Force Experienced = eE
Acceleration $=\frac{\text{eE}}{\text{m}}=\frac{\text{eE}}{\text{Rm}}$
Now $, V^2 = 2 \times a \times s [\because\text{x}=0]$
$\text{V}=\sqrt{\frac{2\text{e}\times\text{V}\times\text{R}}{\text{Rm}}}=\sqrt{\frac{\text{2eV}}{\text{m}}}$
Time taken by particle to cover the arc $=\frac{2\pi\text{m}}{\text{qB}}=\frac{2\pi\text{m}}{\text{eB}}$
Since the acceleration is along ‘Y’ axis.
Hence it travels along x axis in uniform velocity,
Therefore, $\text{v}\times\text{t}=\sqrt{\frac{2\text{em}}{\text{m}}}\times\frac{2\pi\text{m}}{\text{eB}}\sqrt{\frac{8\pi^2\text{mV}}{\text{eB}^2}}$

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