A proton travels through uniform magnetic and electric fields. The magnetic fiel
ID: 1360609 • Letter: A
Question
A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.25 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2940 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 3.72 V/m, (b) in the negative z direction and has a magnitude of 3.72 V/m, and (c) in the positive x direction and has a magnitude of 3.72 V/m?
Explanation / Answer
B = - 2.25i mT, v =2940 j m/s, q =1.6x10-19 C
(a) E =3.72 k V/m
F =q(vxB) +qE
F = 1.6x10-19 [(2940 j)x(-2.25x10-3 i)] +3.72k ]
F =(1.6x10-19)[6.615k +3.72k]
F = 6.201x10-19 N
(b) E = -3.72 k V/m
F =q(vxB) +qE
F = 1.6x10-19 [(2940 j)x(-2.25x10-3 i)] - 3.72k ]
F =(1.6x10-19)[6.615k -3.72k]
F = 4.632x10-19 N
(c) E =3.72 i V/m
F =q(vxB) +qE
F = 1.6x10-19 [(2940 j)x(-2.25x10-3 i)] +3.72i ]
F =(1.6x10-19)[6.615k +3.72i]
F = (1.6x10-19)[(6.615)2 +(3.72)2]1/2
F = 4.55x10-19 N
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