Lorentz force on a proton

Problem

A proton ($q = 1{,}6\cdot 10^{-19}$ C) moves at $v = 3\cdot 10^5$ m/s perpendicular to a uniform magnetic field of magnitude $B = 1$ T. Calculate the magnitude of the Lorentz force.

Solution

The velocity is perpendicular to the field, so $\sin\theta = 1$ and the magnitude of the Lorentz force is $F = |q|\,v\,B = 1{,}6\cdot 10^{-19}\cdot 3\cdot 10^5\cdot 1$ $\ev{F \approx 4{,}8\cdot 10^{-14}\;\text{N}}$

Links

Topics: Magnetismo Concepts: Forza di Lorentz