Lorentz force and radius of a proton

Problem

A proton moves with $v = 2\cdot 10^6$ m/s in a perpendicular field $B = 0{,}5$ T. Calculate (a) the magnitude of the Lorentz force, (b) the radius of the circular orbit.

Solution

(a) The Lorentz force on a charge in a perpendicular field has magnitude: $F = |q| v B = 1{,}6\cdot10^{-19}\cdot 2\cdot10^6 \cdot 0{,}5$ $\ev{F \approx 1{,}6\cdot 10^{-13}\;\text{N}}$

(b) The magnetic force acts as the centripetal force, so $|q|vB = \dfrac{m_p v^2}{r}$, from which: $r = \frac{m_p v}{qB} = \frac{1{,}67\cdot10^{-27}\cdot 2\cdot10^6}{1{,}6\cdot10^{-19}\cdot 0{,}5} \approx 4{,}2\cdot10^{-2}\;\text{m}$ $\ev{r \approx 4{,}2\;\text{cm}}$

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