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}*10^7\text{ m/s}`

To find the pitch.

Step-by-step explanation:

The pitch can be calculated by the formula

`P=(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*3.14*9.1*10^(-31)*3*10^7* cos\text{ 30}^(\circ)}{0.2*1.6*10^(-19)} \\ =\text{ 4.639}*10^(-3)\text{ m} \end{gathered}`