Question

An electron moves a 2.0x10^6 meter per the second perpendicular to a magnetic field having a flux density of 2 teslas. What is the magnitude of the magnetic force on the electron?

Answer 1

Answer:

F_B = 6.4*10^-13 N

Explanation:

The magnetic force on the electron, generated by the motion of the electron and the magnetic field is given by:

$F_B=qv\ X\ B$

q: electron charge = 1.6*10^{-19}C

v: speed of the electron = 2.0*10^6 m/s

B: magnitude of the magnetic field = 2T

However, the direction of B and v are perpendicular between them. So, the angle between vectors is 90°. The magnitude of the magnetic force is:

$|F_B|=qvBsin90\°=qvB$

You replace the values of q, v and B in the last equation:

$|F_B|=(1.6*10^{-19}C)(2.0*10^6m/s)(2T)\\\\|F_B|=6.4*10^{-13}\ N$

hence, the magnetic force on the electron is 6.4*10^-23 N