Question

When a particle of charge q moves with a velocity v⃗ in a magnetic field B⃗ , the particle is acted upon by a force F⃗ exerted by the magnetic field. To find the direction and magnitude of this force, follow the steps in the following Tactics Box. Keep in mind that the right-hand rule for forces shown in step 2 gives the direction of the force on a positive charge. For a negative charge, the force will be in the opposite direction.If the magnetic field of the wire is 4.0×10−4 T and the electron moves at 6.0×106 m/s , what is the magnitude F of the force exerted on the electron?Express your answer in newtons to two significant figures.

Answer 1

Answer:

$3.8\cdot 10^{-16}N$

Explanation:

For a charged particle moving perpendicularly to a magnetic field, the magnitude of the force exerted on the particle is:

$F=qvB$

where

q is the magnitude of the charge of the particle

v is the velocity of the particle

B is the magnetic field strength

In this problem, we have

$q=e=1.6\cdot 10^{-19}$ is the charge of the electron

$v=6.0\cdot 10^6 m/s$ is the velocity

$B=4.0\cdot 10^{-4}T$ is the magnetic field

Substituting into the formula, we find the force:

$F=(1.6\cdot 10^{-19} C)(6.0\cdot 10^6 m/s)(4.0\cdot 10^{-4}T)=3.8\cdot 10^{-16}N$