(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant 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])

3. 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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Introduction to electrodynamics - help with a dipole problem

**Physics Forums | Science Articles, Homework Help, Discussion**