Коллаборативные решения задач по физике
10.1.1. A proton accelerated by a voltage of 20 kV enters a uniform magnetic fieldwith an induction of 0.1 T perpendicular to the field. Find the radius of thecircle along which the proton moves in the magnetic field.
Using:
$$\frac{1}{2}mv^2 = qV, \hspace{0.5cm} r=\frac{mv}{qB}$$
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:$$v= \sqrt{\frac{2qV}{m}} \approx 1.96 \times 10^6 \hspace{0.12cm} \mathrm{m/s}$$$$r \approx 0.2\hspace{0.12cm}\mathrm{m}$$
$$\boxed {\hspace{0.12cm} r \approx 0.2m \hspace{0.12cm} }$$