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 2.85 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1720 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 3.88 V/m, (b) in the negative z direction and has a magnitude of 3.88 V/m, and (c) in the positive x direction and has a magnitude of 3.88 V/m?

Explanation / Answer

here,

magnetic feild , B = - 0.00285 T i

speed of proton , v = 1720 m/s j

a)

electric feild ,E = 3.88 k

the net force , F = e * (E + v X B )

F = 1.6 * 10^-19 * ( 3.88 k + 0.00285 * 1720) k

F = 1.47 * 10^-18 N k

the net force acting is 1.47 * 10^-18 N k

b)

electric feild ,E = - 3.88 k

the net force , F = e * (E + v X B )

F = 1.6 * 10^-19 * ( - 3.88 k + 0.00285 * 1720) k

F = 1.02 * 10^-19 N k

the net force acting is 1.02 * 10^-19 N k

c)

electric feild ,E = 3.88 i

the net force , F = e * (E + v X B )

F = 1.6 * 10^-19 * ( 3.88 i + 0.00285 * 1720 k)

F = 1.6 * 10^-19 * ( 3.88 i + 4.9 k)

|F|= 1.6 * 10^-19 * sqrt(3.88^2 + 4.9^2 )

|F| = 1 * 10^-18 N

the net force acting is 1 * 10^-18 N

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