Magnitude and direction a uniform magnetic field

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SUMMARY

The discussion focuses on determining the magnitude and direction of a uniform magnetic field affecting a proton moving perpendicularly at a velocity of 1.0x107 m/s with an acceleration of 2.0x1013 m/s2. The correct formula used is B = F/qv sin(θ), where θ is 1, q is 1.6x10-19 C, and the mass of the proton is 1.67x10-27 kg. The right-hand rule is emphasized as a crucial tool for visualizing the direction of the magnetic field.

PREREQUISITES
  • Understanding of Lorentz force and magnetic fields
  • Familiarity with the right-hand rule for magnetic forces
  • Basic knowledge of vector mathematics
  • Concept of acceleration in physics
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  • Study the application of the Lorentz force in electromagnetic theory
  • Learn about the right-hand rule and its applications in physics
  • Explore vector cross products and their significance in magnetic fields
  • Investigate the properties of protons and their behavior in magnetic fields
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Physics students, educators, and anyone interested in understanding the dynamics of charged particles in magnetic fields.

physics_ash82
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magnetism

Hi I'm stuck on this problem I its got to be something simple.Here goes...

A proton moves perpendicularly to a uniform magnetic field B at 1.0x10^7 m/s and exhibits an acceleration of 2.0x10^13m/s^2 in the + x-direction when its velocity is in the + z-direction. Determine the magnitude and direction of the field.

I have been trying to use B= F/qvsin(theta)
theta = 1
q = 1.6x10^-19


and then I have the problem :blushing: So any other help would be useful thankx

ash
 
Last edited:
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Think about the use of the right hand rule and think it through. If our not to secure with that and you aregood at vectors math, try working out the cross products and such. If your still stuck let us know what's up.

P.S; the magnitude is given.
 
The right hand rule is a great tool if used. Try to use it to work out the problem, it makes things so much simpler.
 
Thanx

I figured it out I was using the right formula and had forgotten that the mass of the proton is 1.67x10^-27 and I was using the charge which was 1.6 x10^-19.. but thank you for the idea on right hand rule that helped with the 2nd part.

ash:smile:
 

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