Question 7.52: An atomic electron (charge q) circles about the nucleus (cha...

Initially,  \frac{m v^{2}}{r}=\frac{1}{4 \pi \epsilon_{0}} \frac{q Q}{r^{2}} \Rightarrow T=\frac{1}{2} m v^{2}=\frac{1}{2} \frac{1}{4 \pi \epsilon_{0}} \frac{q Q}{r} . After the magnetic field is on, the electron circles in a new orbit, of radius r_{1} \text { and velocity } v_{1}:

\frac{m v_{1}^{2}}{r_{1}}=\frac{1}{4 \pi \epsilon_{0}} \frac{q Q}{r_{1}^{2}}+q v_{1} B \Rightarrow T_{1}=\frac{1}{2} m v_{1}^{2}=\frac{1}{2} \frac{1}{4 \pi \epsilon_{0}} \frac{q Q}{r_{1}}+\frac{1}{2} q v_{1} r_{1} B .

But  r_{1}=r+d r, \text { so }\left(r_{1}\right)^{-1}=r^{-1}\left(1+\frac{d r}{r}\right)^{-1} \cong r^{-1}\left(1-\frac{d r}{r}\right), \text { while } v_{1}=v+d v, B=d B . To first order, then,

T_{1}=\frac{1}{2} \frac{1}{4 \pi \epsilon_{0}} \frac{q Q}{r}\left(1-\frac{d r}{r}\right)+\frac{1}{2} q(v r) d B, \text { and hence } d T=T_{1}-T=\frac{q v r}{2} d B-\frac{1}{2} \frac{1}{4 \pi \epsilon_{0}} \frac{q Q}{r^{2}} d r .

Now, the induced electric field is E=\frac{r}{2} \frac{d B}{d t} (Ex. 7.7), so m \frac{d v}{d t}=q E=\frac{q r}{2} \frac{d B}{d t}, \text { or } m d v=\frac{q r}{2} d B . The increase in kinetic energy is therefore d T=d\left(\frac{1}{2} m v^{2}\right)=m v d v=\frac{q v r}{2} d B . Comparing the two expressions, I conclude that dr = 0. qed