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An electron gun with its collector at a potential of $100 \;V$ fires out electrons in a spherical bulb containing hydrogen gas at low pressure $(10^{-2}\: mm \: of\: Hg)$ A magnetic field of $2.83 \times 10^{-4} \: T$ curves the path of the electrons in a circular orbit of radius $12\; cm$. Find e/m of the electron.

$\begin {array} {1 1} (a)\;0.173 \times 10^8C \ kg & \quad (b)\;1.73 \times 10^{11}C \ kg \\ (c)\;17.3 \times 10^{11}C \ kg & \quad (d)\;17.3 \times 10^8C \ kg \end {array}$

 

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Ans : (b)
So, $ \large\frac{1}{2} $$mv_2 = eV$
$ \Rightarrow v^2= \large\frac{2eV}{m}$………..(i)
So, Centripetal force = magnetic force
$ \Rightarrow \large\frac{mv^2}{r} $$= ev$$B$
$ \Rightarrow eB = \large\frac{mv}{r}$
$ \Rightarrow v = \large\frac{eBr}{m}$………….(ii)
So, putting (ii) in (i),
$\large\frac{2eV}{m} = \large\frac{e^2B^2r^2}{m^2}$
$ \Rightarrow \large\frac{e}{m} = \large\frac{2V}{B^2r^2}$
$= \large\frac{2 \times 100}{[(2.83 \times 10^{-4} )^2 \times (12 \times 10^{-2} )^2 ]} $$= 1.73 \times 10^{11} C / kg$
answered Dec 25, 2013 by thanvigandhi_1
edited Mar 23, 2014 by balaji.thirumalai
 

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