An electron traveling at high (or relativistic) speed moves perpendicularly to
a magnetic field of \(0.20 \mathrm{~T}\). Its path is circular, with a radius of
\(15 \mathrm{~m}\). Find \((a)\) the momentum, \((b)\) the speed, and
(c) the kinetic energy of the electron. Recall that, in nonrelativistic situ
ations, the magnetic force \(q u B\) furnishes the centripetal force \(m u^{2} /
r .\) Thus, since \(p=m v\), it follows that
$$
p=q B r
$$
and this relation holds even when relativistic effects are important.
First find the momentum using \(p=q B r\)
(a) \(p=\left(1.60 \times 10^{-19} \mathrm{C} \times 0.20 \mathrm{~T}\right)(15
\mathrm{~m})=4.8 \times 10^{-19} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\)
(b) Because \(p=m v / \sqrt{1-\left(v^{2} / c^{2}\right)}\) with \(m=9.11 \times
10^{-31}\) kg. we have
$$
4.8 \times 10^{-19} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}=\frac{(m
\mathrm{c})(v / \mathrm{c})}{\sqrt{1-\left(v^{2} / \mathrm{c}^{2}\right)}}
$$
Squaring both sides and solving for \((\mathrm{u} / \mathrm{c})^{2}\) give
$$
\frac{v^{2}}{c^{2}}=\frac{1}{1+3.23 \times 10^{-7}} \quad \text { or } \quad
\frac{v}{c}=\frac{1}{\sqrt{1+3.23 \times 10^{-7}}}
$$
Most hand calculators cannot handle this. Accordingly, we make use of the fact
that \(1 / \sqrt{1+x} \approx 1-\frac{1}{2} x\) for \(x<<1\). Then
$$
\begin{array}{l}
\mathrm{u} / \mathrm{c} \approx 1-1.61 \times 10^{-7}=0.99999984 \\
\text { (c) } \left.\mathrm{KE}=(\gamma m-m) \mathrm{c}^{2}=m \mathrm{c}^{2}
\mid \frac{1}{\sqrt{1-\left(v^{2} / \mathrm{c}^{2}\right)}}-1\right]
\end{array}
$$
But we already found \((\mathrm{u} / \mathrm{c})^{2}=1 /\left(1+3.23 \times
10^{-7}\right)\). If we use the approximation \(1 /(1+x) \approx 1-x\) for
\(x<<1\), we have \((\mathrm{u} / \mathrm{c})^{2} \approx 1-\) \(3.23 \times
10^{-7}\). Then
$$
\mathrm{KE}=m \mathrm{c}^{2}\left(\frac{1}{\sqrt{3.23 \times
10^{-7}}}-1\right)=\left(m \mathrm{c}^{2}\right)\left(1.76 \times
10^{3}\right)
$$
Evaluating the above expression yields
$$
\mathrm{KE}=1.4 \times 10^{-10} \mathrm{~J}=9.0 \times 10^{8} \mathrm{eV}
$$
An alternative solution method would be to use \(\mathrm{E}^{2}=p^{2}
\mathrm{c}^{2}+m^{2} \mathrm{c}^{4}\) and recall that
\(\mathrm{KE}=\mathrm{E}-\mathrm{mc}^{2}\).