Question

An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields are perpendicular to each other, what is the electron's speed?

Answer 1

Answer:

The speed of the electron is $2.55\times 10^3\ m/s$.

Explanation:

Given that,

The magnitude of electric field, $E=1.32\ kV/m=1.32\times 10^3\ V/m$

The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then,  the magnitude of electric field and magnetic field is balanced such that :

$evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s$

or

$v=2.55\times 10^3\ m/s$

So, the speed of the electron is $2.55\times 10^3\ m/s$. Hence, this is the required solution.