A proton travels through uniform magnetic and electric fields. The magnetic fiel

Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.10 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1810 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is:

(a) in the positive z direction and has a magnitude of 3.77 V/m,

(b) in the negative z direction and has a magnitude of 3.77 V/m, and

(c) in the positive x direction and has a magnitude of 3.77 V/m?

Explanation / Answer

magnitude = q v B = 1.60 * 10-19 * 1810 * 3.10 * 10-3 = 8.98 * 10-19 N

a) Electric force = 3.77 * 1.60 * 10-19 = 6.03 * 10-19 N

Net force = (8.98 * 10-19) + (6.03 * 10-19) = 1.50 * 10-18 N

b) Electric force = 3.77 * 1.60 * 10-19 = 6.03 * 10-19 N

Net force = - (6.03 * 10-19) + (8.98 * 10-19) = 2.95 * 10-19 N

c) Electric force = 3.77 * 1.60 * 10-19 = 6.03 * 10-19 N

Net force = 10-19 * sqrt (6.032 + 8.982)  = 10.8 * 10-19 N

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