A proton, traveling with a velocity of 3.4 × 10 6 m/s due east, experiences a ma

Question

A proton, traveling with a velocity of 3.4 × 106 m/s due east, experiences a maximum magnetic force of 8.0 × 10-14 N. The direction of the force is straight down, toward the surface of the earth. What is the magnitude and direction of the magnetic field B, assumed perpendicular to the motion?

A charge of -8.36 C is traveling at a speed of 6.51 × 106 m/s in a region of space where there is a magnetic field. The angle between the velocity of the charge and the field is 55.9o. A force of magnitude8.15 × 10-3 N acts on the charge. What is the magnitude of the magnetic field?

Explanation / Answer

1)

F = qvB

8 x 10^-14 = 1.6 x 10^-19 x 3.4 x 10^6 x B

B = 0.147 T, direction towards you.

2)

F = |q|*v*B*sin

8.15 x 10^-3 = 8.36 x 10^-6 x 6.51 x 10^6 x B x sin(55.9)

B = 1.811 x 10^-4 T

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