An electron moves in the magnetic field B 0590 iT with a sp

An electron moves in the magnetic field B = 0.590 i^T with a speed of 1.40 ×107m/s in the directions shown in the figure. For each, what is magnetic force F on the electron?

Solution

The angle between the velocity vector and magnetic field vector is 45 degrees

F = q v B sin theta

= 1.6 * 10^-19 * 0.590 * 1.40 * 10⁷ * sin 45

= 9.34 * 10^-13 N

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The angle between the velocity vector and magnetic field vector is 90 degrees

F = q v B sin theta

= 1.6 * 10^-19 * 0.590 * 1.40 * 10⁷

= 1.32 * 10^-12 N