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A magnet produces a 0.6000 T field between its poles, directed to the east. A dust particle with charge q = āˆ’4.000 Ɨ 10āˆ’18 C is moving straight down at 0.6000 cm/s in this field. What is the magnitude and direction of the magnetic force on the dust particle? Give your answer in scientific notation.

Respuesta :

Answer:

-1.44 * 10^(-20) N

Explanation:

Parameters given:

Electric Charge, Q = -4 * 10^(-18) C

Velocity, v = 0.6 cm/s = 0.006m/s

Magnetic field, B = 0.6 T

The magnitude of the Magnetic force is given as:

F = Q*v*B

F = -4 * 10^(-18) * 0.006 * 0.6

F = -1.44 * 10^(-20) N

The direction of the force is inward and it is pointed perpendicular to the velocity and magnetic field plane.

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