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Q)

A monoenergetic electron beam with speed of $5.2 \times 10^6\: m/s$ is subjected to a magnetic field of $1.3 \times 10^{-4}T$ normal to the beam velocity. What is the radius of the circle traced by the beam, given $\large\frac{e}{m}$ for electron equals $1.76 \times 10^{11}C /kg$?

$\begin {array} {1 1} (a)\;22.7\: m & \quad (b)\;2.27\: cm \\ (c)\;22.7\: cm & \quad (d)\;2.27\: m \end {array}$

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A)
Ans : (c)
$v=5.2 \times 10^6m/s\: \: \: \: \: \:B= 1.3 \times 10^{-4}T$
So, Force exerted on the electron is : $F = e|v \times B| = evB\: \sin \theta$
$( \theta$=Angle between magnetic field and beam velocity)
So, $\theta=90^{\circ}$
$F=evB$………………..(i)
The beam traces a circular path of radius, $r$.
It is the magnetic field, due to its bending nature, that provides the
centripetal force $(F=\large\frac{mv^2}{r})$ for the beam.
Hence, Eq. (i) reduces to
$evB = \large\frac{mv^2}{r}$
So, $r= \large\frac{mv}{eB} = \large\frac{v}{ \bigg[\bigg(\Large\frac{e}{m}\bigg) B\bigg]}$
$= \large\frac{5.2 \times 10^6 }{(1.76 \times 10^{11} \times 1.3 \times 10^{-4})}$
= 0.227 m = 22.7 cm