A 10 ;eV electron is circulating in a plane a | vedclass

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Question

Kinetic energy of electron $\frac{1}{2} 	imes m v^{2}=10 eV$

and magnetic induction $( B )=10^{-4} Wb / m ^{2}$

Therefore $\frac{1}{2}\left(9.1

Vedclass · Accepted Answer

$11$

Vedclass · Answer

Kinetic energy of electron $\frac{1}{2} 	imes m v^{2}=10 eV$

and magnetic induction $( B )=10^{-4} Wb / m ^{2}$

Therefore $\frac{1}{2}\left(9.1 	imes 10^{-31}ight) v^{2}=10 	imes\left(1.6 	imes 10^{-19}ight)$

$v^{2}=\frac{2 	imes 10 	imes\left(1.6 	imes 10^{-19}ight)}{9.1 	imes 10^{-31}}=3.52 	imes 10^{12}$

$v=1.876 	imes 10^{6} m$.

Centripetal force $=\frac{m v^{2}}{r}=B e v$.

$r=\frac{m v}{B e}=\frac{\left(9.1 	imes 10^{-31}ight) 	imes\left(1.876 	imes 10^{6}ight)}{10^{-4} 	imes\left(1.6 	imes 10^{-19}ight)}$

$=11 	imes 10^{-2} m=11 cm$

A $10 \;eV$ electron is circulating in a plane at right angles to a uniform field at magnetic induction $10^{-4} \;W b / m^{2}(=1.0$ gauss), the orbital radius of electron is ........ $cm$

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