Determine the magnitude and direction of the force on an electron traveling 8.75×10^5 m/s horizontally to the east in a vertically upward magnetic field of strength 0.45 Tesla.
Added by April L.
Step 1
6 x 10^-19 C - Velocity of electron (v) = 8.75 x 10^5 m/s - Magnetic field strength (B) = 0.45 T - Angle between velocity and magnetic field (θ) = 90 degrees Show more…
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