Electromagnetic Force on a Moving Proton: Homework Example

AI Thread Summary
The discussion revolves around solving a homework problem related to the electromagnetic force on a moving proton. The user is confused about the direction of deflection using the right-hand rule and whether work is done when the proton is deflected. Clarifications confirm that the proton will indeed be deflected downwards and that no work is done during this deflection. The user also struggles with calculating the proton's velocity after being accelerated through 10,000V, ultimately determining it to be 1.38 x 10^6 m/s, which leads to a final answer of 7.2 cm for the radius of the proton's path. The conversation highlights the importance of understanding energy changes and motion equations in solving such problems.
Schaus
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Homework Statement


https://lh3.googleusercontent.com/HPLcLqiEmZPOh_7gIM0jNK4cXk6KMgMz2StpKfDC4QDa5Z66Cc-CQ4HKx8nY1n8AF74paz2NM3aaqmVPDVUxAkY9nGhaZGP6f9nu4ItV1cB6GFRASPnt1GI3tmWoOz47r5163k5mJf89FAwKHOkW6XCIeD1pYHAr1J2UXQCPlA1L3njmFTIZOYQYjXpmwO-FQ7WQ3ntnb6kUUpmoxk8UHqYDvxIP04tFgdP6Z654-yOCf-KJaqzbWkv6L4to4kR2BaMA017mxG9ZTVofrA_vuIVVlWV7-9QpguWva93N4GYGItiXDB59hPJpUjVLE6LNKwS04Imvzj1x8NWIswca14eg5KmWe4riQeVybL5sb5xhQ2Q5OZbqMgrS1k-KMtQjoOomr3YkCSEmVWrJSW7ltaW71et85585uwz7VUzGX-mvlro2VklAN349QPfCqqOBr3UUTHOEdaizmutciSU7RH9OMdsyCVEuBuT28icPDkmeC5rDRjnwEaUMbCX5ntOSCL9hrPkhEEpMY7nF1vee4vNoPIfZjVRa4UP5l7kGj_RNIcqlTqdORWIscWpCFypHTRZRurdZ8b_MLPr2lVZBY0TLEC4MK_kC7oIgE_C41_phTjGW3dQC=w757-h450-no

UP5l7kGj_RNIcqlTqdORWIscWpCFypHTRZRurdZ8b_MLPr2lVZBY0TLEC4MK_kC7oIgE_C41_phTjGW3dQC=w757-h450-no.png


Homework Equations


F = QvB
V = kQ/r

The Attempt at a Solution


I'm just learning this stuff and I wasn't given an example of any kind so I don't really know how to start this question...
For 8.a) If I'm using the RHR and my thumb is facing in the x direction then I believe the proton will be deflected downwards? I was also under the impression that work done had to be parallel to the force and if the proton is being deflected then there is no work done?

8.b) I tried doing mv/r = qB and rearranging the formula for r = mv/qB but if I don't have velocity and I'm left with 2 variables I'm at a loss.

Any help would be greatly appreciated.
 
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Schaus said:
For 8.a) If I'm using the RHR and my thumb is facing in the x direction then I believe the proton will be deflected downwards?
Correct.

I was also under the impression that work done had to be parallel to the force and if the proton is being deflected then there is no work done?
Correct

8.b) I tried doing mv/r = qB and rearranging the formula for r = mv/qB but if I don't have velocity and I'm left with 2 variables I'm at a loss.
What is the proton's velocity after it is accelerated through 10,000V? I think you should assume that it starts from rest.
 
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If I assume its zero then my radius would be zero? Or would I have to use a kinematics or energy formula to solve? If so I don't know what else I have to go on to use the formulas or which formula.
 
I meant that you should assume that the proton's velocity is zero before it is accelerated through 10,000V. Do you know how to calculate its velocity after it has been accelerated through 10,000 V, assuming that it is at rest before the acceleration starts? How much energy will it gain after it has been accelerated through 10,000 V?
 
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Would I use the formula ΔV=ΔEp/Q. Otherwise I'm not sure how to find the speed with voltage.
 
You are correct that the change in energy is Q ΔV. Then what is the velocity?
 
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I found the velocity to be 1.38 x 106m/s. I plugged that into the rest of the info I had and got my answer of 7.2cm! Thanks for your help!
 
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