![\"current](\"http:\/\/pages.vassar.edu\/magnes\/files\/2014\/04\/current-wire-B-only.jpg\")
<\/a><\/p>\nThis system has a net charge of zero, so there should be no electric field. However, the sum of the charges does cause a total current <\/span>I = <\/i>2<\/span>\u03bbv<\/i> to the right, which gives a magnetic field of B = \u03bc<\/span>0<\/sub>\u03bb<\/span>v<\/i>\/\u03c0s in the \u03d5 direction (as dictated by the right-hand rule), for any distance s from the center of the wire.<\/span><\/p>\n Now, consider the same situation, but in the reference frame where q is at rest. Suddenly, the positive linear charge is much smaller than the negative one, leaving a net negative charge on the wire, which will produce an electric field. Simply by changing the reference frame, the situation switched from a purely magnetic phenomenon to a combination of electric and magnetic.<\/p>\n However, rather than trying to analyze the magnitude of the new charge and current values to find E and B, it is possible to simply use the following transformation equations to find out what the new E and B fields are.<\/p>\n $E’_x$ = $E_x$ Where $\\gamma$ = $\\frac{1}{\\sqrt{1-{\\frac{v^2}{c^2}}}}}$ (From Griffiths Introduction to Electrodynamics, 3rd Ed.<\/em>\u00a0<\/em>p531).\u00a0The only hiccup is that the original setup is in cylindrical coordinates, while the transformation equations are in Cartesian. However, Mathematica can do this automatically, so I\u2019m going to leave it in these terms.<\/p>\n B-field:<\/a><\/p>\n<\/a><\/p>\n
\n$E’_y$ = $\\gamma$ ($E_y$ – $vB_z$)
\n$E’_z$ = $\\gamma$ ($E_z$ – $vB_y$)
\n$B’_x$ = $B_x$
\n$B’_y$ = $\\gamma$ ($B_y$ + $\\frac{v}{c^2}$ $E_z$)
\n$B’_z$ = $\\gamma$ ($B_z$ – $\\frac{v}{c^2}$ $E_y$)<\/p>\n<\/a> E-field:<\/a><\/p>\n