Magnetic Force in a Current Carrying Wire

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SUMMARY

The discussion focuses on calculating the magnetic force on electrons within a current-carrying wire. The magnetic field strength at a distance r from the wire's axis is given by the formula B = μ₀I/2πr. The force on an electron is derived using F = QvB, leading to the conclusion that the correct expression for the force is F = (1/2)μ₀v²e²nr, where e is the fundamental charge, n is the electron density, and v is the velocity of the electrons. The key to solving the problem lies in correctly substituting the current I using the current density J.

PREREQUISITES
  • Understanding of electromagnetic theory, specifically the Lorentz force law.
  • Familiarity with the concepts of current density and its relation to electron flow.
  • Knowledge of the fundamental charge (e) and its significance in electromagnetic calculations.
  • Ability to perform integration in the context of physics problems.
NEXT STEPS
  • Study the derivation of the current density J = nev in detail.
  • Learn about the integration process for calculating current in cylindrical coordinates.
  • Explore the implications of the Lorentz force in different geometries of current-carrying conductors.
  • Investigate the applications of magnetic fields in practical scenarios, such as in electromagnets and motors.
USEFUL FOR

Students of physics, particularly those studying electromagnetism, as well as educators and anyone involved in teaching or learning about the behavior of currents in magnetic fields.

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Homework Statement



The current in a wire consists of n electrons per unit volume moving with a velocity v. What is the effect on these electrons of the magnetic field set up by the current itself within the wire? What is the force on one electron at a distance r from the centre of the wire?


Homework Equations



F = QvBsinθ
Magnetic field strength at a distance r from the axis of the wire outside the wire = u0I/2∏r

The Attempt at a Solution



I know that for part 1, the effect of the magnetic field on the electrons inside the wire will be a radially outward force (taking all electrons to be traveling in straight lines along the wire).

Part 2

F will be equal to QvBsin90 = QvB

Making the subsitution for B I have

F = Qvu0I/2∏r

I believe now that I have to make subsitution for I. Which would be 4/3∏a3vne

where e is fundamental charge and a is the radius of the wire.

This gives:

F = (3/8)u0ve2na3r

I know that the answer is:

F = (1/2)u0v2e2nr
I think I'm making a mistake in calculating my current, but I'm not sure. Help Please?

Thanks
 
Last edited:
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Hi! This is my first time helping someone out so here goes:

To make the substitution for current I, you should use:
I=\int J \bullet da
where J is the current density, equal to nev in this case. After doing a quick integration you should get the proper current which will give the correct answer for the force.
 

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