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A positively charged particle of mass 7.21×10-8kg is traveling due east with a speed of 89.9m/s and enters a 0.290T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 2.06×10-3s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. What is the magnitude of the magnetic force acting on the particle?

Work done so far:

F = ma = mv^2/r

F = B(q x v)sin90

Travels 1/4 of the circle so I figured the distance, X is (2pi*r)/4.

In order to find F(mag), need to find the two unknowns: q and r. So i figured I would try to find r first.

I used X = vt + 1/2at^2, substituted a with a = v^2/r, with X = (2pi*r)/4.

I ended up getting an unsolvable algebraic expression in terms of r as the unknown.

I know all I have to do, once r is determined is solve for q. Then I can calculate F from F = B(v x q)sin90.

Any suggestions??

Thanks!

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# Homework Help: Magnetic Forces

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