Calculate the force on a particle from magnetic field of infinite wire

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SUMMARY

The discussion centers on calculating the magnetic force on a charged particle moving near an infinitely long wire carrying a current, denoted as "i". The magnetic field strength, "B", is derived using the formula B = (μ₀i)/(2πr). The magnitude of the force, "F", is expressed as F = (q*v*μ₀*i*sin(θ))/(2πr). The participants confirm that while the magnetic field is non-uniform, the derived equation for force at a specific radius and angle is accurate.

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Homework Statement


Problem: Determine an expression for the magnitude of the magnetic force on a charged particle moving near an infinitely long wire, carrying a current i.
  • Particle with charge q
  • Magnitude of the particles velocity = |v|
  • Magnetic field strength = B
  • Current = i

Homework Equations


B can be calculated with:
B=\frac{μ_{0}i}{2\pi{r}}

Magnitude of force can be calculated with:
F=q*v*b*sin(θ) (Cross product of the particles velocity vector and the magnetic field vector)

The Attempt at a Solution


I've combined these equations to come up with a single equation for the magnitude of the force on the particle:

F=\frac{q*v*μ_{0}*i*sin(θ)}{2\pi{r}}

However this final equation is the part I am not confident on. I've proceeded on the basis the the magnetic field around an infinite wire carrying current is non-uniform; that is the direction of the magnetic field is not constant, but would appreciate my working being checked!

Thanks in advance!
 
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The magnetic field is not constant, this will change sin(θ) and r in a trajectory. But the equation for the force at a specific radius with a specific angle is correct.
Current is written with a capital I. "i" is usually the imaginary unit.
 


Looks good.

BTW i = √(-1) is only for physicists.

Real men (engineers) use j = √(-1) and i is very regularly used to denote current :smile:
 


Physics Forums > Science Education > Homework & Coursework Questions > Introductory Physics >

rude man said:
BTW i = √(-1) is only for physicists.
And mathematics as well.
 


mfb said:
And mathematics as well.

Yeah, but not real men (and women)! :smile:
 
Thanks both! Just needed that extra pair of eyes to look at it :smile:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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