A particle of charge +2.00*10 -6 C has a velocity of 2.00*10 7 m/s in the direct

Question

A particle of charge +2.00*10-6C has a velocity of 2.00*107m/s in the directions described below. This particle moves in a magnetic field which is constant and uniform 0.500T and points in the positive y direction.

Find the magnitude and direction of the force acting on the particle due to the magnetic field for the following situations.

1. The velocity is in the negative y direction (down). F = ?

2. The direction of the Force from "part 1" above is.... a) up (+y) b) down (-y) c) left (-x) d) right (+x) e) into the page (-z) f) out of the page (+z) g) can't be determined or no direction.

If you could explain how you came up with each answer, I would be most appreciative!

Explanation / Answer

I assume the charged perticle moving +x direction.

1)magnetic force on moving charged particle, F = q*v*B*sin(theta)

= 2*10^-6*2*10^7*0.5*sin(90)

= 20 N

2) Direction : towards +z

To find the direction of magnetic force use right hand rule.

Right hand Rule : take your fingures in the diretion of motion of charge and culr towards magnetic filed, then thum indicates the direction of

magnetic force.

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