Deriving the Force on an Infinitessimal current-loop in a Magnetic Field

Click For Summary
SUMMARY

The discussion focuses on deriving the equation for the force on an infinitesimal current loop in a magnetic field, represented as F = ∇(m·B). The derivation involves evaluating the integral F = ∫ I(dℓ × B) along a square loop with sides of length ε, positioned parallel to the yz plane. The user successfully applies Taylor expansion to express B(0,y,ε) and B(0,ε,z) in terms of B(0,y,0) and B(0,0,z), leading to the formulation of the force components. The solution manual's approximation, ∫dz(∂B/∂y)|_{0,0,z} ≈ ε(∂B/∂y)|_{0,0,0}, is questioned for its validity.

PREREQUISITES
  • Understanding of vector calculus, specifically the cross product and gradient operations.
  • Familiarity with magnetic fields and the Biot-Savart law.
  • Proficiency in Taylor series expansion and its application in physics.
  • Knowledge of LaTeX for formatting mathematical expressions.
NEXT STEPS
  • Study the Biot-Savart law and its implications for magnetic fields generated by current loops.
  • Learn about Taylor series expansion in the context of physics and its applications in approximating functions.
  • Explore vector calculus techniques, particularly the gradient and divergence operations in electromagnetism.
  • Review LaTeX syntax for mathematical expressions to improve clarity in scientific communication.
USEFUL FOR

Physicists, electrical engineers, and students studying electromagnetism who seek to understand the derivation of forces on current loops in magnetic fields.

OGrowli
Messages
12
Reaction score
0
We're asked to derive the following equation:

[tex]F=\triangledown(\vec{m}\cdot \vec{B})[/tex]

by evaluating [tex]F=\int I(d\vec{l}\times \vec{B})[/tex] along a sqaure loop with sides of length ε, parralel to the yz plane. The square's bottom left corner is situated at the origin.

so far I have,
[itex]d\vec{l} \times \vec{B}= -dz \times \hat{z} \times \vec{B}(0,0,z) + dy \hat{y} \times \vec{B}(0,y,0)+dz\hat{z} \times \vec{B}(0,\epsilon ,z)-dy \hat{y} \times \vec{B}(0,y,\epsilon)[/itex]

then using taylor expansion to turn B(0,y,ε) and B(0,ε,z) into expressions of B(0,y,0) and B(0,0,z) respectively and combining like terms I've got:

[itex] - \hat{y} \times \left \epsilon dy\frac{\partial \vec{B}} {\partial z} \right |_{0,y,0} + \hat {z} \times \left \epsilon dz \frac{\partial \vec{B}} {\partial y} \right |_{0,0,z} [/itex]

the next step would be to evaluate ∫dF and the solutions manual says that,

[itex] \left \int dz\frac{\partial\vec{B}}{\partial y} \right |_{0,0,z}\approx \left \epsilon \frac{\partial\vec{B}}{\partial y} \right |_{0,0,0}[/itex]

from this point the solution comes pretty easily, but I still don't understand why the above is true. Can anyone explain this to me?
 
Physics news on Phys.org
OGrowli said:
[itex] - \hat{y} \times \left \epsilon dy\frac{\partial \vec{B}} {\partial z} \right |_{0,y,0} + \hat {z} \times \left \epsilon dz \frac{\partial \vec{B}} {\partial y} \right |_{0,0,z} [/itex]

[itex] \left \int dz\frac{\partial\vec{B}}{\partial y} \right |_{0,0,z}\approx \left \epsilon \frac{\partial\vec{B}}{\partial y} \right |_{0,0,0}[/itex]

Does anyone know what is wrong with the LaTex coding here?
 

Similar threads

Replies
3
Views
2K
Replies
2
Views
1K
Replies
3
Views
3K
  • · Replies 25 ·
Replies
25
Views
2K
Replies
12
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
64
Views
6K