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Question-53156




Question Number 53156 by peter frank last updated on 18/Jan/19
Answered by peter frank last updated on 18/Jan/19
eV=(1/2)mv^2   (e/m)=(v^2 /(2V))
$${eV}=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \\ $$$$\frac{{e}}{{m}}=\frac{{v}^{\mathrm{2}} }{\mathrm{2}{V}} \\ $$$$ \\ $$
Answered by peter frank last updated on 18/Jan/19
ii     eV=(1/2)mv^2 ...... (i)  v=(E/B)............(ii)[F_(e ) =F_(m ) =eE=Bev  sub eqn ii in i  (e/m)=(E^2 /(2VB^2 ))  E=(v/d)  (e/m)=(E^2 /(2VB^2 ))  k=(v/B^2 )  kB^2 =v  B^2 α v
$${ii}\:\:\:\:\:{eV}=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ……\:\left({i}\right) \\ $$$${v}=\frac{{E}}{{B}}…………\left({ii}\right)\left[{F}_{{e}\:} ={F}_{{m}\:} ={eE}={Bev}\right. \\ $$$${sub}\:{eqn}\:{ii}\:{in}\:{i} \\ $$$$\frac{{e}}{{m}}=\frac{{E}^{\mathrm{2}} }{\mathrm{2}{VB}^{\mathrm{2}} } \\ $$$${E}=\frac{{v}}{{d}} \\ $$$$\frac{{e}}{{m}}=\frac{{E}^{\mathrm{2}} }{\mathrm{2}{VB}^{\mathrm{2}} } \\ $$$${k}=\frac{{v}}{{B}^{\mathrm{2}} } \\ $$$${kB}^{\mathrm{2}} ={v} \\ $$$${B}^{\mathrm{2}} \alpha\:{v} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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