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Question

Electrons (e,m) emitted with negligible speed from an electron gun are accelerated through a potential difference V0 along the x-axis. These electrons emerge from a narrow hole into a uniform magnetic field of strength B directed along x-axis. Some electrons emerging at slightly divergent angles as shown. These electrons are refocused on the x -axis for second time at a distance Nπ2mV0eB2. Find the value of N ___.


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Solution

From the figure, the electrons will be refocussed after one time period of revolution for their helical path.

Thus, they will cover a distance equal to pitch, till they are refocussed.

Let, d= pitch,
v= speed of electron along x axis
K= Kinetic energy of electron

d=vx×T

d=v(2πmeB) ........(1)

Here,

v=2Km=2eV0m .....(2)

From equations (1) & (2);

d=2eV0m×2πmeB

d=(2πmeB)22eV0m

d=8π2mV0eB2

Particles meet for second time at a distance 2d=32π2mV0eB2

Comparing this with the data given in the question we get, N=32
Why this question ?
Tip: When electrons are accelerated through a potential difference of V0 Volt, they will gain K.E equal to loss in P.E, i.e.,
12mv2=eV0

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