Exercise 4.1.3: The field of a single moving charge

$R=R{a}_y$, so $v\times R=vR(a_x\times a_y)=vR{a}_z$ and

$R=R{a}_x\parallel v$, so $v\times R=0$: $B=0$ straight ahead of (and behind) the charge. The field lines are circles round the line of motion, strongest in the plane through the charge perpendicular to $v$ (Figure 4.1.4).

At $(0,R,0)$, $E=\frac{e}{4\pi {\epsilon }_0{R}^2}=9\times 10^9\times \frac{1.602\times 10^{-19}}{10^{-18}}=1.44\times 10^9$ V/m, so $B/E=1.6\times 10^{-2}/(1.44\times 10^9)=1.11\times 10^{-11}$ s/m. In general, with $v\perp R$,

since ${\mu }_0{\epsilon }_0=1/c^2$ (with the constants of this book, ${\mu }_0{\epsilon }_0=4\pi \times 10^{-7}\times 10^{-9}/36\pi =1/(9\times 10^{16})$ s${}^2$/m${}^2$ exactly, i.e. $c=3\times 10^8$ m/s). Here $v/c^2=10^6/(3\times 10^8{)}^2=1.11\times 10^{-11}$ s/m, as found.

Units: ${\mu }_{0}ev/R^2$ is (H/m)(C)(m/s)/m${}^2$ $=$ (H/m)(A/m) $=$ T. The magnetic force between two charges moving side by side is $v^2/c^2$ times their electric force, $10^{-5}$ here. Magnetism is a tiny relativistic correction to electricity, which becomes visible in a wire only because the wire is electrically neutral: the enormous electric forces cancel and the small magnetic one is all that is left. Problem 4.1.1 is this formula summed over $nqvd{v}^{\prime }$ in a neutral wire.