Calculate the force on an electron that enters a uniform magnetic field of flux density 150 mT at a speed of 8.0 x 10^6 m/s in a perpendicular direction to the field.*1 point1.0 x 10^-13 N1.9 x 10^-13 N1.0 x 10^-12 N1.9 x 10^-12 N
Question
Calculate the force on an electron that enters a uniform magnetic field of flux density 150 mT at a speed of 8.0 x 10^6 m/s in a perpendicular direction to the field.*1 point1.0 x 10^-13 N1.9 x 10^-13 N1.0 x 10^-12 N1.9 x 10^-12 N
Solution
The force on a charged particle moving in a magnetic field is given by the equation F = qvBsinθ, where:
F is the force, q is the charge of the particle, v is the speed of the particle, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.
In this case, the electron is moving perpendicular to the field, so θ = 90 degrees and sinθ = 1.
The charge of an electron (q) is -1.6 x 10^-19 C, the speed (v) is 8.0 x 10^6 m/s, and the magnetic field strength (B) is 150 mT, which is equivalent to 150 x 10^-3 T.
Substituting these values into the equation gives:
F = (-1.6 x 10^-19 C) * (8.0 x 10^6 m/s) * (150 x 10^-3 T) * 1 F = -1.92 x 10^-13 N
The force is negative because the electron is negatively charged, but the magnitude of the force is 1.92 x 10^-13 N.
So, the closest answer to the options given would be 1.9 x 10^-13 N.
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