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April 17th, 2011, 12:51 AM   #1
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Moving proton, Magnetic field, velocity, force

Here for another question! (you guys are great by the way thank you).
Okay so, this is the question

A proton moving in a uniform magnetic field with v1= 1.30*10^6 m/s experiences force F1 = 1.59*10^-16 k_hat N. A second proton with
v2 = 2.49*10^6 j_hat m/s experiences F2 = -4.03*10^-16 k_hat N in the same field.

What is the magnitude of B? What is the direction of B? Give your answer as an angle measure ccw from the +x-axis.

Now, I know how to go about solving this. F=qv x B. I get the right magnitude of values, being:
B1 = 7.644*10^-4 j_hat T
B2=1.012*10^-3 i_hat T

Thus, B=1.27*10^-3 T or 1.27mT

However, in determining the direction of B1 and B2, I get negative values, when then should be positive apparently. I'm sure I'm using the right hand rule for force correctly.. but it doesnt seem to be working.
I attached the drawings I did to help me out. The correct angle is supposed to be 37.1
Attached Images
File Type: png Screen shot 2011-04-17 at 1.46.31 AM.png (50.4 KB, 2452 views)
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April 17th, 2011, 12:58 PM   #2
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Re: Moving proton, Magnetic field, velocity, force

Whoops, I put in cw from +x-axis instead of ccw from the + x-axis. Fixed it in the problem and posted this in case no one replied because that made it not make sense...
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April 18th, 2011, 12:25 AM   #3
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Re: Moving proton, Magnetic field, velocity, force

I see now that if I flip my and j_hat coordinates, the right hand rule gives positive values for both B1 and B2. This still doesnt make much sense to me though... it seems like it could be either both positive or both negative. In either case B is still in the same place on a coordinate axis, it's just facing in opposite directions...

Still doesn't apply to the angle/direction part though. It definitely exists 37.1 above the +x-axis, but is that the direction??..hmm
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October 27th, 2011, 07:39 AM   #4
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Re: Moving proton, Magnetic field, velocity, force

Not sure if you still need an answer, but here goes. You found B1 and B2, in the j and i hat direction respectively. It is simply vector addition to find the angle. You can set these up on the xy-plane and use the inverse tangent function to find the angle (Hint draw the two vectors as a triangle on xy-plane). It's not the only way to find the angle so if this is confusing just let me know!
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