From the period to the field

Problem

A proton cyclotron has orbital period $T = 6{,}5\cdot 10^{-8}$ s. Find the magnitude of $\vec B$ using $T = 2\pi m/(qB)$.

Solution

The period of circular motion in a magnetic field is $T = \dfrac{2\pi m}{qB}$, independent of the velocity. Inverting for $B$: $B = \frac{2\pi m_p}{qT} = \frac{2\pi\cdot 1{,}67\cdot10^{-27}}{1{,}6\cdot10^{-19}\cdot 6{,}5\cdot10^{-8}}$ $\ev{B \approx 1{,}0\;\text{T}}$

Links

Topics: Magnetismo Concepts: Moto circolare uniforme