An electron having kinetic energy of 100 eV c | vedclass

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(A) The kinetic energy of the electron is $K = 100 eV = 100 	imes 1.6 	imes 10^{-19} J = 1.6 	imes 10^{-17} J$.The radius of the circular path is $

Vedclass · Accepted Answer

$3.3 	imes 10^{-4} T$

Vedclass · Answer

(A) The kinetic energy of the electron is $K = 100 eV = 100 	imes 1.6 	imes 10^{-19} J = 1.6 	imes 10^{-17} J$.The radius of the circular path is $r = 10 cm = 0.1 m$.The mass of the electron is $m = 0.5 MeV/c^2 = \frac{0.5 	imes 10^6 	imes 1.6 	imes 10^{-19} J}{(3 	imes 10^8 m/s)^2} \approx 8.89 	imes 10^{-31} kg$.The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.Rearranging for $B$,we get $B = \frac{\sqrt{2mK}}{rq}$.Substituting the values: $B = \frac{\sqrt{2 	imes 8.89 	imes 10^{-31} 	imes 1.6 	imes 10^{-17}}}{0.1 	imes 1.6 	imes 10^{-19}}$.$B = \frac{\sqrt{28.448 	imes 10^{-48}}}{1.6 	imes 10^{-20}} = \frac{5.33 	imes 10^{-24}}{1.6 	imes 10^{-20}} \approx 3.33 	imes 10^{-4} T$.

An electron having kinetic energy of $100 eV$ circulates in a path of radius $10 cm$ in a magnetic field. The magnitude of magnetic field $|B|$ is approximately [Mass of electron $= 0.5 MeV/c^2$,where $c$ is the velocity of light].

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