# Calculus and Vectors

The magnetic force (Vector FM)on a particle in a magnetic field is found by Vector FM = Vector I × Vector B, Vector I is the charge multiplied by the velocity of a charged particle and Vector B the strength of the magnetic field, in Tesla (T). An electron accelerates from rest to the right, in a horizontally directed electric field. The electron then leaves the electric field at a speed of 4.0 × 106, entering a magnetic field of magnitude 0.20 T (Tesla) directed into the screen. Calculate the magnitude and direction of the magnetic force on the electron. (Charge on an electron: q = 1.6 × 10-19)

I can't figure this one out guys any solutions?

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1. I assume we are using the common xyz (i-j-k) coordinates where x points out to the front of the screen. That means
I = qv = (1.6*10^-19)(4.0*10^6) j = 6.4*10^-13 j
B = -0.20 i
Now, you know that i×j = k
That means j×-i = k
F = 6.4*10^-13 j × -0.20 i = 1.28*10^-13 k

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oobleck

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