A proton, traveling with a velocity of 2.0 10 6 m/s due east, experiences a maxi

Question

A proton, traveling with a velocity of 2.0 106 m/s due east, experiences a maximum magnetic force of 8.0 10-14 N. The direction of the force is straight down, toward the surface of the earth. What is the magnitude and direction of the magnetic field B, assumed perpendicular to the motion?

A proton, traveling with a velocity of 2.0 10^6 m/s due east, experiences a maximum magnetic force of 8.0 10^-14 N. The direction of the force is straight down, toward the surface of the earth. What is the magnitude and direction of the magnetic field B, assumed perpendicular to the motion? B = ?T West or North or South or East

Explanation / Answer

Fmax = q*v*B....

B = Fmaax/(q*v)   = (8*10^-14)/(1.6*10^-19*2*10^6) = 0.25 T along south

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