Introduction to electrodynamics - help with a dipole problem

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SUMMARY

This discussion focuses on solving a dipole problem in electrodynamics, specifically analyzing the magnetic field \(\vec{B}\) generated by a dipole moment \(\vec{m} = m\hat{k}\) located at the origin. Key equations include \(\vec{B} = \frac{μ_{0}m}{4∏r^3}(2\cosθ\hat{r} + \sinθ\hat{θ})\) and its coordinate-free form \(\vec{B} = \frac{μ_{0}}{4∏r^3}[3(\vec{m}\bullet\hat{r})\hat{r} - \vec{m}]\). The discussion addresses the angles between the magnetic field and the z-axis at various positions, including the z-axis, y-axis, and a cone at θ=45 degrees. Participants provide insights into the calculations and transformations necessary to derive the magnetic field components.

PREREQUISITES
  • Understanding of magnetic dipoles and their properties
  • Familiarity with vector calculus and coordinate systems
  • Knowledge of the Biot-Savart law and magnetic field equations
  • Proficiency in trigonometric functions and their applications in physics
NEXT STEPS
  • Study the derivation of the magnetic field from a dipole using the Biot-Savart law
  • Learn about the implications of the angle θ in magnetic field calculations
  • Explore the concept of magnetic field lines and their visualization
  • Investigate the effects of varying dipole moments on the magnetic field distribution
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism, as well as researchers and professionals working with magnetic fields and dipole interactions.

Flynndle
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Homework Statement


Let the dipole [itex]\vec{m}[/itex] = m[itex]\hat{k}[/itex] be at the origin, and call a certain horizontal axis the y axis.
a) On the z axis, what is the angle between the z axis and the magnetic field?
b) On the y axis, what is the angle between the z axis and the magnetic field?
c) On the cone θ=45 degrees, what is the angle between the z axis and the magnetic field?
d) What is the angle of the cone on which the magnetic field is horizontal?

Homework Equations


I believe there is some relevance to the equation: [itex]\vec{B}[/itex] =[itex]\frac{μ_{0}m}{4∏r^3}[/itex](2cosθ[itex]\hat{r}[/itex]+sinθ[itex]\hat{θ}[/itex])


The Attempt at a Solution


I tried putting the previous equation into the coordinate free form to try if that would help.

[itex]\vec{B}[/itex] =[itex]\frac{μ_{0}}{4∏r^3}[/itex][3([itex]\vec{m}[/itex][itex]\bullet\hat{r}[/itex])[itex]\hat{r}[/itex]-[itex]\vec{m}[/itex]]

I then simplified this to:


[itex]\vec{B}[/itex] =[itex]\frac{μ_{0}}{4∏r^3}[/itex]m[3cosθ-1][itex]\hat{k}[/itex]

I was not sure what to do after this so I tried:

r^2=x^2+y^2+z^2

set x=0 so it's in the yz plane,

r=(y^2+z^2)^(1/2)

arccos(z/r)=θ

arccos[itex]\frac{z}{(y^2+z^2)^(1/2)}[/itex] = θ
 
Last edited:
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The position vector is written as
[itex]\vec r = x \hat i+y\hat j +z\hat k[/itex].
If [itex]\vec r[/itex] encloses the angle θ with the z axis and the angle φ with the positive x axis, x=r sinθ cosφ, y=r sinθ sinφ, and z=rcosθ. The unit vector along [itex]\vec r[/itex] is
[itex]\hat r = \sin(\theta)\cos(\phi)\hat i+\sin(\theta)\sin(\phi)\hat j +\cos(\theta)\hat k[/itex].

Use all of these to get [itex]\vec B[/itex].

[itex](\vec m\cdot \hat r)=m\cos(\theta)[/itex], and it is multiplied by [itex]\hat r[/itex], so [itex]\vec B[/itex] has x, y, and z components. Try to find it.

ehild
 

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