Question Number 38376 by ajfour last updated on 24/Jun/18
Commented by MrW3 last updated on 25/Jun/18
$${what}'{s}\:{the}\:{question}? \\ $$
Commented by ajfour last updated on 25/Jun/18
$${position}\:{vector}\:{of}\:{the}\:{particle} \\ $$$${as}\:{a}\:{function}\:{of}\:{time}\:,\:\:\:\bar {{r}}\left({t}\right)=? \\ $$$${Consider}\:{the}\:{current}\:{carrying} \\ $$$${wire}\:{to}\:{be}\:{very}\:{long}\:{and}\:{gravity} \\ $$$${present}. \\ $$
Answered by ajfour last updated on 25/Jun/18
$${Magnetic}\:{field}\:{at}\:\left({x},{y}\right)\:{is} \\ $$$$\:\:\:\:{B}=−\frac{\mu_{\mathrm{0}} {I}}{\mathrm{2}\pi{x}}\hat {{k}} \\ $$$${let}\:{velocity}\:{at}\:{P}\left({x},{y}\right)\:{be} \\ $$$$\:\:\:\:\:\:\bar {{v}}={v}_{{x}} \hat {{i}}+{v}_{{y}} \hat {{j}} \\ $$$$\bar {{F}}\:=\:−{mg}\hat {{j}}+\frac{\mu_{\mathrm{0}} {Iq}}{\mathrm{2}\pi{x}}\left(−{v}_{{y}} \hat {{i}}+{v}_{{x}} \hat {{j}}\right) \\ $$$${m}\left(\frac{{dv}_{{x}} }{{dt}}\hat {{i}}+\frac{{dv}_{{y}} }{{dt}}\hat {{j}}\right)=−{mg}\hat {{j}}+\frac{\mu_{\mathrm{0}} {Iq}}{\mathrm{2}\pi{x}}\left(−{v}_{{y}} \hat {{i}}+{v}_{{x}} \hat {{j}}\right) \\ $$$$\frac{{dv}_{{x}} }{{dt}}\:=\:−\frac{\mu_{\mathrm{0}} {Iqv}_{{y}} }{\mathrm{2}\pi{x}} \\ $$$$\frac{{dv}_{{y}} }{{dt}}\:=\:−{mg}+\frac{\mu_{\mathrm{0}} {Iqv}_{{x}} }{\mathrm{2}\pi{x}} \\ $$$$….\:? \\ $$