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Montano1993 [528]
3 years ago
11

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 t

he magnitude of the magnetic force on the electron?
Physics
1 answer:
Darina [25.2K]3 years ago
6 0

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

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