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Explain Bot To find the magnitude of the force experienced by a moving charged particle in a magnetic field, you can use the equation:
F = q * v * B * sin(theta)
where:
F is the magnitude of the force
q is the charge of the particle
v is the velocity of the particle
B is the strength of the magnetic field
theta is the angle between the velocity vector and the magnetic field vector.
Let's use this equation to solve the given problems:
1. Magnitude of the force experienced by a proton:
Given:
Velocity (v) = 3 km/s = 3,000 m/s
Magnetic field strength (B) = 0.05 T
Since the problem doesn't mention the angle between the velocity and the magnetic field, we can assume that they are perpendicular to each other (theta = 90 degrees). In this case, sin(theta) = 1.
The charge of a proton (q) is +1.6 x 10^-19 C (coulombs).
Using the equation, we can calculate the magnitude of the force:
F = (1.6 x 10^-19 C) * (3,000 m/s) * (0.05 T) * (1)
F = 2.4 x 10^-14 N
Therefore, the magnitude of the force experienced by the proton is 2.4 x 10^-14 Newtons.
2. Magnitude of the magnetic force on an electron:
Given:
Velocity (v) = 2 x 10^5 m/s
Magnetic field strength (B) = 1.4 x 10^-3 T
Angle between the velocity and the magnetic field (theta) = 45 degrees
Using the equation, we need to calculate the sine of 45 degrees (sin(45)) which is √2 / 2.
The charge of an electron (q) is -1.6 x 10^-19 C (coulombs).
Using the equation, we can calculate the magnitude of the force:
F = (-1.6 x 10^-19 C) * (2 x 10^5 m/s) * (1.4 x 10^-3 T) * (√2 / 2)
F = -1.792 x 10^-13 N
Therefore, the magnitude of the force experienced by the electron is 1.792 x 10^-13 Newtons.
Remember that the negative sign indicates that the force is in the opposite direction of the magnetic field.