Confusion in the derivation of the force on a magnetic dipole

Click For Summary

Homework Help Overview

The discussion revolves around the derivation of the force on a magnetic dipole, specifically the expression \( f = \nabla (\vec{m} \cdot \vec{B}) \). Participants are examining two different proofs that arrive at this result, one assuming the curl of the magnetic field is zero and the other not making this assumption.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the differences between two proofs regarding the force on a magnetic dipole, questioning the validity of assumptions made in each proof. They discuss the implications of the curl of the magnetic field being zero versus non-zero and how this affects the derivation.

Discussion Status

Some participants express confusion about the implications of Griffiths' proof and whether it can be reconciled with the first proof. There is an ongoing exploration of how to continue the first proof without assuming the curl is zero, with some guidance provided on potential manipulations and identities that could be used.

Contextual Notes

Participants note that the magnetic field in question is due to external sources, which may influence the curl at the location of the dipole. There is also mention of specific examples from Griffiths' text that illustrate scenarios where the curl is non-zero.

  • #31
wait i was looking through wikipedia and found this
f5d9854a6668509bf50f8d6db476f2c16895ef0c

this seems to suggest
##
(A \cdot \nabla)B
##
is not
##
A \cdot (\nabla B)
##
 
Physics news on Phys.org
  • #32
timetraveller123 said:
wait i was looking through wikipedia and found this
f5d9854a6668509bf50f8d6db476f2c16895ef0c

this seems to suggest
##
(A \cdot \nabla)B
##
is not
##
A \cdot (\nabla B)
##
Here, the notation ##\nabla B## is not the dyadic formed from ##\nabla## and ##B##. So, here,
##A \cdot (\nabla B) \neq (A \cdot \nabla)B##.

Here, ##\nabla B## is the gradient of the vector B as define at https://en.wikipedia.org/wiki/Gradient#Gradient_of_a_vector
or in (1.14.3) here http://homepages.engineering.auckla...ensors/Vectors_Tensors_14_Tensor_Calculus.pdf

In matrix notation, the gradient of B, ##\nabla B##, as used in the Wikipedia article is the transpose of the matrix representing the dyadic ##\left( \nabla B \right)_{\rm dyadic}##.

See also the thread https://www.physicsforums.com/threads/gradients-of-vectors-and-dyadic-products.139000/
 
  • Like
Likes   Reactions: timetraveller123
  • #33
ok thanks again i will look more into the subject
 

Similar threads

Replies
3
Views
2K
Replies
2
Views
1K
  • · Replies 25 ·
Replies
25
Views
2K
Replies
1
Views
2K
Replies
19
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
2K