Question

Question:A 2.50 MeV electron moves perpendicularly to a magnetic field in a path with a 3.0 cm radius of curvature. What is the magnetic field B? (See Problem 53)

Short answer

The magnetic field for the electron is $0.178T$.

Step-by-step solution

Identification of given data

The energy of electron is $E=2.50MeV$

The radius of curvature is $r=3cm$

The magnetic field for the electron is found by equating the necessary centripetal force by magnetic force on the electron.

Determination of magnetic field for electron

The magnetic field for electron is given as:

$B=\frac{\sqrt{2mE}}{qr}$

Here, $q$ is the charge of electron and its value is $1.6\times {10}^{-19}C$ , m is the mass of electron and its value is $9.1\times {10}^{-31}kg$

Substitute all the values in the above equation.

$$\begin{gathered} B=\frac{\sqrt{2\left(9.1\times {10}^{-31}kg\right)\left(2.50MeV\right)\left({\displaystyle \frac{1.6\times {10}^{-13}J}{1MeV}}\right)}}{\left(1.6\times {10}^{-19}C\right)\left(3cm\right)\left({\displaystyle \frac{{10}^{-2}m}{1cm}}\right)} \\ B=0.178t \end{gathered}$$

Therefore, the magnetic field for the electron is $0.178T.$