Question

An electron enters a uniform magnetic field 0.20 T at an angle of 30° to the field. Determine the pitch of the helical path assuming its speed is 3×107 ms-1

Answer 1

Given:

The magnetic field is B = 0.2 T

The angle is

$\theta\text{ = 30}^{\circ}$

The speed of the helical path is

$v\text{ = 3}\times10^7\text{ m/s}$

To find the pitch.

Explanation:

The pitch can be calculated by the formula

$P=\frac{2\pi m}{Bq}vcos\theta$

Here, m is the mass of the electron whose value is 9.1 x 10^(-31) kg

q is the charge of the electron whose value is 1.6 x 10^(-19) C

On substituting the values, the pitch will be

$\begin{gathered} P=\text{ }\frac{2\times3.14\times9.1\times10^{-31}\times3\times10^7\times cos\text{ 30}^{\circ}}{0.2\times1.6\times10^{-19}} \\ =\text{ 4.639}\times10^{-3}\text{ m} \end{gathered}$