1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
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

  1. Dec 11, 2011 #1
    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
     
    Last edited: Dec 11, 2011
  2. jcsd
  3. Dec 13, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Introduction to electrodynamics - help with a dipole problem
Loading...