10.1.1 | revision #14448 | Savchenko Solutions

Statement

10.1.1. A proton accelerated by a voltage of 20 kV enters a uniform magnetic field
with an induction of 0.1 T perpendicular to the field. Find the radius of the
circle along which the proton moves in the magnetic field.

Solution

Using:

$\frac{1}{2}mv^2 = q, \hspace{0.5cm} r=\frac{mv}{qB}$

With:
$q=1.6 \times 10^{-19}C, \hspace{0.5cm} m=1.67 \times 10^{-27}kg, \hspace{0.5cm} V=20kV, \hspace{0.5cm} B=0.10T$

We get:
$v= \sqrt{\frac{2qV}{m}} \approx 1.96 \times 10^6 \, \mathrm{m/s}$
$r \approx 0.2\mathrm{m}$

Answer

$\boxed { r \approx 0.2m }$