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Question-53686




Question Number 53686 by ajfour last updated on 25/Jan/19
Commented by ajfour last updated on 25/Jan/19
Charged particle (q,m) released  at origin. Find velocity v^�  as a  function of the z coordinate.  Electric field, E_0 k^�  ,   Magnetic field, −B_0  j^�  are present.
$${Charged}\:{particle}\:\left({q},{m}\right)\:{released} \\ $$$${at}\:{origin}.\:{Find}\:{velocity}\:\bar {{v}}\:{as}\:{a} \\ $$$${function}\:{of}\:{the}\:{z}\:{coordinate}. \\ $$$${Electric}\:{field},\:{E}_{\mathrm{0}} \hat {{k}}\:,\: \\ $$$${Magnetic}\:{field},\:−{B}_{\mathrm{0}} \:\hat {{j}}\:{are}\:{present}. \\ $$
Commented by Tinkutara last updated on 25/Jan/19
v_z =(√(2Aωz−ω^2 z^2 ));  A=(E_0 /B_0 )  ω=((qB_0 )/m)
$${v}_{{z}} =\sqrt{\mathrm{2}{A}\omega{z}−\omega^{\mathrm{2}} {z}^{\mathrm{2}} }; \\ $$$${A}=\frac{{E}_{\mathrm{0}} }{{B}_{\mathrm{0}} } \\ $$$$\omega=\frac{{qB}_{\mathrm{0}} }{{m}} \\ $$
Commented by ajfour last updated on 25/Jan/19
seems correct,v_x =?
$${seems}\:{correct},{v}_{{x}} =? \\ $$
Commented by Tinkutara last updated on 25/Jan/19
v_x (t)=(E_0 /B_0 )(1−cos ωt)
$${v}_{{x}} \left({t}\right)=\frac{{E}_{\mathrm{0}} }{{B}_{\mathrm{0}} }\left(\mathrm{1}−\mathrm{cos}\:\omega{t}\right) \\ $$
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].
$${not}\:{right}\:{process},\:{sir};\:{soon}\:{v}_{{x}} \\ $$$${develops}\:,\:{so}\:{you}\:{must}\:{take}\:{this} \\ $$$${into}\:{consideration},\:{moreover}, \\ $$$$\bar {{v}}\:\:{is}\:{asked}\:{for}\:{as}\:{a}\:{function}\:{of}\:{z}, \\ $$$${and}\:{not}\:{time}\:{t}\left[\:\bar {{v}}\left({t}\right)\:{might}\:{be}\:{still}\right. \\ $$$$\left.{more}\:{difficult}\right]. \\ $$

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