Question Number 53342 by aseerimad last updated on 20/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19
$${pls}\:{clarify}\:{direction}\:{of}\: \\ $$$$\overset{\rightarrow} {{E}}={electric}\:{field}\:\:\left({is}\:{it}\:+{ve}\:{x}\:{axis}\:\right) \\ $$$$\overset{\rightarrow} {{V}}={velocity}\:{proton}\left({is}\:{it}\:+{ve}\:{x}\:{axis}\right. \\ $$$$\overset{\rightarrow} {{B}}={magnectic}\:{induction}\:\left({is}\:{it}\:−{z}\:{axis}\right) \\ $$$${Lorentz}\:{force} \\ $$$$\overset{\rightarrow} {{F}}={q}\overset{\rightarrow} {{E}}+\overset{\rightarrow} {{V}}×\overset{\rightarrow} {{B}} \\ $$$${after}\:{clarification}... \\ $$$$\overset{\rightarrow} {{F}}_{{E}} ={q}_{{proton}} ×\left({E}\overset{\rightarrow} {{i}}\right)\:{force}\:{due}\:{to}\:{electric}\:{field} \\ $$$$\overset{\rightarrow} {{F}}_{{E}} ={qE}\overset{\rightarrow} {{i}} \\ $$$$\overset{\rightarrow} {{F}}_{{m}} ={q}\left\{{V}\overset{\rightarrow} {{i}}×\left(−{B}\overset{\rightarrow} {{k}}\right)\right\}=\left(−\mathrm{1}\right){qVB}\left(\overset{\rightarrow} {{i}}×\overset{\rightarrow} {{k}}\right) \\ $$$$=\left(−\mathrm{1}\right){qVB}\left(−{j}\right) \\ $$$$={qVB}\left({j}\right) \\ $$$${electric}\left[{force}\:{along}\:+{ve}\:{x}\:{axis}\right. \\ $$$${magnetic}\:{force}\:+{ve}\:{y}\:{axis} \\ $$$${so}\:{proton}\:{move}\:{in}\:{x}−{y}\:{plane} \\ $$$${net}\:{force}=\left({qE}\right){i}+\left({qVB}\right){j} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\overset{\rightarrow} {{S}}={ix}+{jy}\:\leftarrow{path}\:{way} \\ $$$$\:\:\:\:\:\:\:\:\:={i}\left({u}_{{x}} {t}+\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{{qE}}{{m}}\right){t}^{\mathrm{2}} \right)+{j}\left({u}_{{y}} {t}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{qVB}}{{m}}\right){t}^{\mathrm{2}} \right. \\ $$$$\:\:\:\:\:\:\:\:\:\:={i}\left\{{Vt}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{qE}}{{m}}\right){t}^{\mathrm{2}} \right\}+{j}\left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{qVB}}{{m}}\right){t}^{\mathrm{2}} \right\} \\ $$$$\overset{\rightarrow} {\:}\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by aseerimad last updated on 21/Jan/19
$${srry}.\:\:\:\:{velocity}\:{of}\:{proton}\:{alongx}\:{axis}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{electric}\:{field}\:{along}\:{x}\:{axis}\:{and} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{magnetic}\:{field}\:{along}\:−{z}\:{axis} \\ $$
Commented by aseerimad last updated on 21/Jan/19
May the Almighty reward you with goodness!
Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19
$${blessing}\:{shower}\:{to}\:{all}... \\ $$