A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of...
A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.10 107 m/s and experiences an acceleration of 1.80 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.
magnitude =
Homework Answers
acceleration = net force / mass
= qVB / m
where q = charge
V = velocity
B = magnetic field
m = mass of the proton
1.8*10^13 = 1.6*10^-19*2.1*10^7*B / 1.67*10^-27
B = 0.00894 T
= 8.94 mT
magntiude of magnetic field = 8.94 mT
direction is -ve y-direction using right hand rule
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