An electron experiences 1.2 x 10^-3 force when it enters the external magnetic field, B with a velocity v. What is the force experienced by the electron if the magnetic field is increased two times and the velocity is decreased to half?
Any help would be greatly appreciated
F=qvBsinα
F1=q(v/2)(2B)sinα = qvBsinα = 1.2x10^-3 T
Thanks so much Elena
To calculate the force experienced by the electron, we can use the formula for the force on a charged particle moving through a magnetic field:
F = q * v * B * sin(θ)
Where:
F is the force experienced by the electron
q is the charge of the electron (which is -1.6 x 10^-19 C)
v is the velocity of the electron
B is the magnetic field
θ is the angle between the velocity vector and the magnetic field vector (which is 90 degrees or π/2 radians if the electron's velocity is perpendicular to the magnetic field)
Given that the electron experiences a force of 1.2 x 10^-3 when entering the magnetic field, we can rewrite the formula as:
1.2 x 10^-3 = (-1.6 x 10^-19) * v * B * sin(θ)
We are asked to find the force experienced by the electron when the magnetic field is increased by a factor of 2 and the velocity is decreased by a factor of 1/2. Let's denote the new force as F2, the new magnetic field as 2B, and the new velocity as v/2.
Using the same formula, we have:
F2 = (-1.6 x 10^-19) * (v/2) * (2B) * sin(θ)
To compare the forces, we can divide F2 by the original force:
F2 / 1.2 x 10^-3 = (-1.6 x 10^-19) * (v/2) * (2B) * sin(θ) / (-1.6 x 10^-19) * v * B * sin(θ)
Canceling out common factors, we have:
F2 / 1.2 x 10^-3 = (v/2) * (2B) * sin(θ) / (v * B * sin(θ))
Simplifying further:
F2 / 1.2 x 10^-3 = (v * 2B * sin(θ)) / (2 * v * B * sin(θ))
The sin(θ) term cancels out:
F2 / 1.2 x 10^-3 = 2
Therefore, F2 = 2 * 1.2 x 10^-3
F2 = 2.4 x 10^-3
So the force experienced by the electron, when the magnetic field is increased two times and the velocity is decreased to half, is 2.4 x 10^-3 units.