Chapter 27 Problem 28

Problem

An electron enters a uniform magnetic field $B=0.28{\text{ T }}$ at a 45° angle to ${\vec {B}}$. Determine the radius $r$ and pitch $p$ (distance between loops) of the electron’s helical path assuming its speed is $3.0\times 10^{6}{\text{m}}{\big /}{\text{s}}$.

Solution

The centripetal force is caused by the magnetic field

$F=qvB\sin \theta =qv_{\perp }B=m{\frac {v_{\perp }^{2}}{r}}$ $\rightarrow r={\frac {mv_{\perp }}{qB}}={\frac {\left(9.11\times 10^{-31}{\text{ kg}}\right)\left(3.0\times 10^{6}{\text{ m}}{\big /}{\text{s}}\right)\sin 45^{\circ }}{\left(1.60\times 10^{-19}{\text{ C}}\right)\left(0.28{\text{ T}}\right)}}=4.314\times 10^{-5}{\text{ m}}$

The component of the velocity that is parallel to the magnetic field is unchanged, and so the pitch is that velocity component times the period of the circular motion.

$T={\frac {2\pi r}{v_{\perp }}}={\frac {2\pi {\frac {mv_{\perp }}{qB}}}{v_{\perp }}}={\frac {2\pi m}{qB}}$

$P=v_{\parallel }T=v\cos 45^{\circ }\left({\frac {2\pi m}{qB}}\right)$ $=\left(3.0\times 10^{6}{\text{ m}}{\big /}{\text{s}}\right)\cos 45^{\circ }{\frac {\left(9.11\times 10^{-31}{\text{ kg}}\right)}{\left(1.60\times 10^{-19}{\text{ C}}\right)\left(0.28{\text{ T}}\right)}}=2.7\times 10^{-4}{\text{ m}}$