Magnitude of force on a Dust Particle

[crazuiee]

Homework Statement

A magnet produces a 0.40 T field between its poles, directed horizontally to the east. A dust particle with charge q = -8.0×10-18 C is moving vertically downwards with a speed of 0.30 cm/s in this field. Whilst it is in the magnetic field, the dust particle is also in an electric field of strength 1.00×10-2 V/m pointing to the north.

(a) What is the magnitude and direction of the magnetic force on the dust particle?
(b) What is the magnitude and direction of the electric force on the dust particle?
(c) What is the magnitude of the net force on the dust particle?

Homework Equations

FB= qvBsinθ
E= FE/q

The Attempt at a Solution

For part a) i got 9.6x10^-21N to the north according to the right hand rule
For part b) i got a magnitude of 4.44x10^-3N but I'm not sure how to find the direction of the force.

 

[kuruman]

What is the direction of the force due to the electric field? Draw a diagram of the two forces, then add them as vectors to find the resultant.

 

[kerimek]

For part c) I would use the Pythagorean theorem to find the magnitude of the net force, which would be 9.6x10^-21N to the north.



Your attempts at solving the problem are correct. To find the direction of the electric force, we can use the right hand rule again. Since the electric field is pointing to the north and the charge on the dust particle is negative, the electric force will be in the opposite direction, to the south. Therefore, the net force on the dust particle will be directed to the north, with a magnitude of 9.6x10^-21N, as you have correctly calculated. This net force will cause the dust particle to move in a curved path, with a radius determined by the balance between the magnetic and electric forces. This is known as the Lorentz force and is a fundamental concept in the study of electromagnetism. Keep up the good work!