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 3.46 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1930 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 5.02 V/m, (b) in the negative z direction and has a magnitude of 5.02 V/m, and (c) in the positive x direction and has a magnitude of 5.02 V/m? Units TN (a) Numbe 1.87e-18 a (b) Number 2.65e-19 Units TN (c) Number Un

Explanation / Answer

B = 3.46 i mT

v = 1930 j m/s

a) f = q(E+V*B) = 1.6*10^-19(5.02K-6.68K) = 2.656 *10^-19 k N(MAGNITUDE)

b) f = 1.6*10^-19(-5.02-6.68) = 18.72 *10^-19 k N (MAGNITUDE)

c) f = 1.6*10^-19(5.02i-6.68k) = 13.37*10^-19 N (MAGNITUDE)

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