An electron moves at 2 x 10^2 m/s perpendicular to a magnetic field of 2T. What is the magnitude of magnetic force?
Correct answer: B. 6.4 x 10^-17N
- A. 1 x 10^-6N
- B. 6.4 x 10^-17N
- C. 6.4 x 10^17N
- D. 4 x 10^6N
Explanation
To calculate the magnitude of the magnetic force experienced by an electron moving perpendicular to a magnetic field, you can use the formula for the magnetic force on a charged particle: F = |q| * v * B Where: F is the magnitude of the magnetic force (in Newtons, N). |q| is the absolute value of the charge of the electron (which is approximately 1.602 x 10^-19 C, the elementary charge of an electron). v is the magnitude of the velocity of the electron (in meters per second, m/s). B is the magnitude of the magnetic field (in Tesla, T). Given: v = 2 x 10^2 m/s B = 2 T |q| ≈ 1.602 x 10^-19 C Now, plug in the values: F = |q| * v * B F = (1.602 x 10^-19 C) * (2 x 10^2 m/s) * (2 T) F ≈ 6.408 x 10^-17 N After calculating the expression, the magnitude of the magnetic force on the electron is approximately 6.408 x 10^-17 N.
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