How Does Torque Relate to Magnetic Dipole Moment in Physics?

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SUMMARY

The discussion centers on deriving the relationship between torque and magnetic dipole moment using the Lorentz force law, specifically the equation Fmag = ∫ I (d× B). Participants clarify that the magnetic dipole moment, denoted as m, is defined as m = Ia, where I is the current and a is the area vector. The conversation highlights the need for a clear understanding of the derivative step in the torque equation to successfully demonstrate that torque is expressed as m × B in a uniform magnetic field.

PREREQUISITES
  • Understanding of the Lorentz force law and its application in electromagnetism.
  • Familiarity with magnetic dipole moment and its calculation.
  • Knowledge of vector calculus, particularly cross products.
  • Access to Griffiths' "Introduction to Electrodynamics" (4th edition) for reference.
NEXT STEPS
  • Study the derivation of torque from the Lorentz force law in detail.
  • Explore the concept of magnetic dipole moment in various current configurations.
  • Learn about the implications of torque in uniform magnetic fields.
  • Review vector calculus techniques, focusing on cross products and their applications in physics.
USEFUL FOR

Students of physics, particularly those studying electromagnetism, educators teaching these concepts, and anyone interested in the mathematical foundations of torque and magnetic dipole moments.

ghoul
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Homework Statement


Starting from the Lorentz force law, in the form of Fmag = ∫ I (d× B), show that the torque on any steady current distribution (not just a square loop) in a uniform field B is m × B.

Homework Equations


I looked at the solution I found in this thread here, but I don't understand the step after you take the derivative of both sides of the torque equation. Why would you use that method in the next step, and how? Where did it come from?

I have Griffiths 4ed.

The Attempt at a Solution


This is all I know how to do.
ubsqgCZ.jpg
 
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ghoul said:

Homework Statement


Starting from the Lorentz force law, in the form of Fmag = ∫ I (d× B), show that the torque on any steady current distribution (not just a square loop) in a uniform field B is m × B.

Homework Equations


I looked at the solution I found in this thread here, but I don't understand the step after you take the derivative of both sides of the torque equation. Why would you use that method in the next step, and how? Where did it come from?

I have Griffiths 4ed.

The Attempt at a Solution


This is all I know how to do.
ubsqgCZ.jpg

Hi mister ghoul! :oldsmile:

What is ##\mathbf m##?
 
Last edited:
I like Serena said:
i mister ghoul! :oldsmile:

What is ##\mathbf m##?
Hi,
I believe that m is magnetic dipole moment. m = Ia
 
ghoul said:
Hi,
I believe that m is magnetic dipole moment. m = Ia

Okay... so what is ##\mathbf a##?
We'll need some expression to substitute, which will hopefully lead to the requested result.
 

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