Question Number 53686 by ajfour last updated on 25/Jan/19
Commented by ajfour last updated on 25/Jan/19
$${Charged}\:{particle}\:\left({q},{m}\right)\:{released} \\ $$$${at}\:{origin}.\:{Find}\:{velocity}\:\bar {{v}}\:{as}\:{a} \\ $$$${function}\:{of}\:{the}\:{z}\:{coordinate}. \\ $$$${Electric}\:{field},\:{E}_{\mathrm{0}} \hat {{k}}\:,\: \\ $$$${Magnetic}\:{field},\:−{B}_{\mathrm{0}} \:\hat {{j}}\:{are}\:{present}. \\ $$
Commented by Tinkutara last updated on 25/Jan/19
$${v}_{{z}} =\sqrt{\mathrm{2}{A}\omega{z}−\omega^{\mathrm{2}} {z}^{\mathrm{2}} }; \\ $$$${A}=\frac{{E}_{\mathrm{0}} }{{B}_{\mathrm{0}} } \\ $$$$\omega=\frac{{qB}_{\mathrm{0}} }{{m}} \\ $$
Commented by ajfour last updated on 25/Jan/19
$${seems}\:{correct},{v}_{{x}} =? \\ $$
Commented by Tinkutara last updated on 25/Jan/19
$${v}_{{x}} \left({t}\right)=\frac{{E}_{\mathrm{0}} }{{B}_{\mathrm{0}} }\left(\mathrm{1}−\mathrm{cos}\:\omega{t}\right) \\ $$
Commented by ajfour last updated on 25/Jan/19
![not right process, sir; soon v_x develops , so you must take this into consideration, moreover, v^ is asked for as a function of z, and not time t[ v^ (t) might be still more difficult].](Q53788.png)
$${not}\:{right}\:{process},\:{sir};\:{soon}\:{v}_{{x}} \\ $$$${develops}\:,\:{so}\:{you}\:{must}\:{take}\:{this} \\ $$$${into}\:{consideration},\:{moreover}, \\ $$$$\bar {{v}}\:\:{is}\:{asked}\:{for}\:{as}\:{a}\:{function}\:{of}\:{z}, \\ $$$${and}\:{not}\:{time}\:{t}\left[\:\bar {{v}}\left({t}\right)\:{might}\:{be}\:{still}\right. \\ $$$$\left.{more}\:{difficult}\right]. \\ $$