Radius and period of an electron

Problem

An electron ($m_e = 9{,}11\cdot 10^{-31}$ kg, $e = 1{,}60\cdot 10^{-19}$ C) enters perpendicularly into a uniform magnetic field $B = 1{,}0$ mT with velocity $v = 4{,}0\cdot 10^5$ m/s. Calculate the radius and period of the circular motion, and draw the direction of rotation knowing that $\vec B$ points out of the page.

Solution

In circular motion the magnetic force acts as the centripetal force: $evB = \dfrac{m_e v^2}{r}$, from which the radius

$r = \frac{m_e v}{eB} = \frac{9{,}11\cdot10^{-31}\cdot 4{,}0\cdot10^5}{1{,}6\cdot10^{-19}\cdot 1{,}0\cdot10^{-3}} \approx 2{,}3\cdot10^{-3}\;\text{m}$

$\ev{r \approx 2{,}3\;\text{mm}}$

The period does not depend on the velocity:

$T = \frac{2\pi r}{v} = \frac{2\pi m_e}{eB} = \frac{2\pi\cdot 9{,}11\cdot10^{-31}}{1{,}6\cdot10^{-19}\cdot 1{,}0\cdot10^{-3}} \approx 3{,}6\cdot10^{-8}\;\text{s}$

$\ev{T \approx 36\;\text{ns}}$

For a negative charge (the electron) in a field $\vec B$ pointing out of the page, the direction of rotation is CLOCKWISE.

Links

Topics: Magnetism Concepts: Lorentz force · Uniform circular motion