Radius of a proton in the field

Problem

A proton enters perpendicularly into a uniform magnetic field of $0{,}50$ T with velocity $2{,}7\cdot 10^5$ m/s. Calculate the radius of the circular trajectory. ($m_p = 1{,}67\cdot 10^{-27}$ kg)

Solution

The magnetic force acts as the centripetal force: $qvB = \dfrac{m_p v^2}{r}$, from which

$r = \frac{m_p v}{qB} = \frac{1{,}67\cdot10^{-27}\cdot 2{,}7\cdot10^5}{1{,}6\cdot10^{-19}\cdot 0{,}50}$

$\ev{r \approx 5{,}7\cdot 10^{-3}\;\text{m}}$

Links

Topics: Magnetismo Concepts: Forza di Lorentz · Moto circolare uniforme