Question

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?

Answer 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
<b>I</b> = q<b>v</b> = (1.6*10^-19)(4.0*10^6) <b>j</b> = 6.4*10^-13 <b>j</b>
<b>B</b> = -0.20 <b>i</b>
Now, you know that <b>i</b>×<b>j</b> = <b>k</b>
That means <b>j</b>×-<b>i</b> = <b>k</b>
<b>F</b> = 6.4*10^-13 <b>j</b> × -0.20 <b>i</b> = 1.28*10^-13 <b>k</b>