Answer to Question #183730 in Electricity and Magnetism for ary

Question

An electron experiences the greatest force as it travels 2.9 × 106 m s-1

in a magnetic 

field when it is moving northward. The force is upward and of 7.2 × 10-13 N 

magnitude. What are the magnitude and direction of the magnetic field? How to 

reduce deflection force on the electron?

Answer 1

$v=2.9 \cdot 10^6 \; m/s; \; F =7.2 \cdot 10^{-13} N.$

The Lorentz force acting on the electron is

$F = qvB sin \alpha$

where $\alpha$ is the angle between magnetic field and velocity. The greatest force, given in the task, is possible when $sin \alpha =1$ which happens when particle velocity is perpendicular to the magnetic field.

$\displaystyle B = \frac{F}{ev} = \frac{7.2 \cdot 10^{-13}}{1.6 \cdot 10^{-19}\cdot 2.9 \cdot 10^6}= 1.552\; T.$

To reduce the force on the electron, we need to either decrease the magnetic field or change the angle between electron direction and magnetic field.