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 2.25 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1660 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 4.64 V/m, (b) in the negative z direction and has a magnitude of 4.64 V/m, and (c) in the positive x direction and has a magnitude of 4.64 V/m?

Explanation / Answer

a)
here

F = q * E + q * V x B

F = q * ( E + V x B)

F = (1.602 * 10^-19) * ( (4.64 k) + 1660 j * (-2.25 * 10^-3) i ) ....................( j * i = -k)

F = 1.602 * 10^-19 * ( 4.64 k + 3.735 k )

F = 7.43 * 10^-19 k + 5.98 * 10^-19 k

F = (13.41 * 10^-19 )k

b)

Fb does not change but Fe does

F = (- 7.43 * 10^-19) k + (5.98 * 10^-19) k

F = -1.45 * 10^-19 k ( in negative z direction)

c)

F = ( 7.43 * 10^-19 ) i + ( 5.98 * 10^-19) k

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