Magnitude of force on a Dust Particle

AI Thread Summary
The discussion focuses on calculating the forces acting on a dust particle in a magnetic and electric field. The magnetic force on the dust particle, calculated using the formula FB = qvBsinθ, is determined to be 9.6x10^-21 N directed to the north. The electric force is calculated to have a magnitude of 4.44x10^-3 N, but the direction remains unclear to the participants. A suggestion is made to draw a diagram to visualize the forces and find the resultant vector. The discussion emphasizes the importance of understanding both the magnitude and direction of forces in physics problems.
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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.
 
Physics news on Phys.org
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.
 

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