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Quantum_Grid

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## Homework Statement

A proton is moving through a vertical magnetic field. The (instantaneous) velocity of the proton is 8.5x10

^{5}m/s horizontally in the north direction. The (instantaneous) acceleration produced by the magnetic force is 2.90x10

^{5}m/s

^{2}in the west direction. What is the magnitude of the magnetic field?

## Homework Equations

Proton Mass = 1.67x10

^{-23}kg

Proton Charge = +e = 1.60x10

^{-19}C

B = F/(qv) (I think!)

## The Attempt at a Solution

Okay, this seems like it should be simple, but the online homework keeps saying I am wrong, so obviously I am skipping something here. First, applying Newton's second law, I found the force, F, required of the Magnetic Force to produce the acceleration (F=ma = (1.67x10

^{-23}kg x 2.90x10

^{5}m/s

^{2}= 4.843x10

^{-18}N). Then, all I need to do is put this force in the equation for a charged particle in motion in a magnetic field, which I think is B=F/(qv), so I should get, B = (4.843x10

^{-18}N)/[(1.60x10

^{-19}C)(8.5x10

^{5}m/s), which gives me B = 3.561x10

^{-5}Tesla.

But apparently that is wrong. Am I missing something small, like a number, or do I have the wrong equation?

Thanks!

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