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?
Okay so I know there's the formula
F = BIlsin0
But isn't the question missing some information? I'm very confused
F=qvBsinα
F1=q(v/2)(2B)sinα = qvBsinα = 1.2x10^-3 T
Yes, you are correct. The formula you mentioned, F = BIl sinθ, calculates the magnetic force experienced by a charged particle moving through a magnetic field. However, the given question is indeed missing some information. In order to calculate the force experienced by the electron after the changes in magnetic field and velocity, we would need to know the current (I) and the angle (θ) between the velocity vector and the magnetic field vector.
Without this additional information, it is not possible to determine the answer. It is important to have the complete information in order to accurately calculate the force.
Yes, you're correct! The formula you mentioned, F = BILsinθ, represents the magnetic force experienced by a charged particle moving in a magnetic field. In this formula, B represents the magnetic field strength, I represents the current (in this case, we can consider it as the electron's charge), L represents the length of the object, θ represents the angle between the direction of the magnetic field and the direction of the current, and sinθ is the sine of that angle.
In the given scenario, we are not provided with the length of the object or the angle θ. However, we can still solve the problem by considering the other given information.
Let's denote the original force experienced by the electron as F1, and its original velocity as v1. The force experienced by the electron when the magnetic field is increased two times and the velocity is decreased to half can be denoted as F2.
According to the given information, F1 = 1.2 x 10^-3 (original force)
If the magnetic field is increased two times, we can represent the new magnetic field as 2B. Similarly, if the velocity is decreased to half, the new velocity can be represented as v1/2.
Now, we need to find F2, the new force experienced by the electron in the altered conditions.
Using the formula F = BILsinθ, we can compare the original and the altered conditions:
F1 = B1ILsinθ1
F2 = (2B)(I)(L)(sinθ2)
Since the electron is the same, the charge, I, remains constant in both conditions. Therefore, we can write:
F1 = F2
B1ILsinθ1 = (2B)(I)(L)(sinθ2)
B1sinθ1 = 2Bsinθ2
Since sinθ is a ratio of two sides of a right triangle, it is constant for the same angle. Therefore, we can conclude that sinθ1 = sinθ2.
Now we have:
B1 = 2B
Since B1 = F1 / (IL) and B = F2 / (IL), we can write:
F1 / (IL) = 2(F2 / (IL))
This simplifies to:
F1 = 2F2
Therefore, the new force experienced by the electron, F2, is half of the original force, F1.