Chapter 20, Problem 87GP

To determine

The radius and pitch of the electron’s helical path.

Answer to Problem 87GP

The radius of the electron’s helical path is $5.25\times {10}^{-5}\text{m}$ .

The pitch of the electron’s helical path is $3.29\times {10}^{-4}\text{m}$

Explanation of Solution

Given:

The magnetic field is $B=0.23\text{T}$

The initial velocity of electron is $3\times {10}^6\text{m}/\text{s}$ .

The initial angle is $\theta =45°$ .

Formula used:

The centripetal force on the proton is equal to the magnetic force on it is,

  $\begin{array}{l}\frac{m_e{\left(v\sin \theta \right)}^2}{r}=Bev\sin \theta \\ \frac{m_{e}v\sin \theta }{r}=Be\\ r=\frac{m_{e}v\sin \theta }{Be}\end{array}$

Here, $B$ is the magnetic field, e is the charge on the proton, v is the speed of the proton, and $v\cos \theta$ is the component of velocity along the magnetic field, $v\sin \theta$ is the perpendicular to the field.

The expression for the pitch of the electron is,

  $p=2\pi r$

Here, r is the radius of the helical path.

Calculation:

The radius of the electron’s helical path is,

  $\begin{array}{c}r=\frac{m_{e}v\sin \theta }{Be}\\ =\frac{\left(9.11\times {10}^{-31}\right)\left(3\times {10}^6\right)\sin 45°}{0.23\times 1.6\times {10}^{-19}}\\ =5.25\times {10}^{-5}\text{m}\end{array}$

The pitch of the electron’s helical path is,

  $\begin{array}{c}p=2\pi r\\ =2\pi \left(5.25\times {10}^{-5}\right)\\ =3.29\times {10}^{-4}\text{m}\end{array}$

Conclusion:

The radius of the electron’s helical path is $5.25\times {10}^{-5}\text{m}$ .

The pitch of the electron’s helical path is $3.29\times {10}^{-4}\text{m}$