Introduction to electrodynamics - help with a dipole problem

In summary, the position vector and unit vector can be used to calculate the magnetic field (\vec B) at various points with a given dipole moment (\vec m) at the origin. The equation for \vec B can be simplified using trigonometric functions and the position vector, and can be used to find the angle between the z axis and the magnetic field at different points such as on the z axis, y axis, and a cone with a certain angle (θ).
  • #1
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
 
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  • #2
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
 

FAQ: Introduction to electrodynamics - help with a dipole problem

1. What is an electric dipole?

An electric dipole is a pair of equal and opposite charges separated by a small distance. This creates a dipole moment, which is a measure of the strength and orientation of the dipole.

2. How does an electric dipole interact with an external electric field?

An electric dipole will experience a torque when placed in an external electric field. This torque causes the dipole to align with the electric field, with the positive charge facing towards the negative end of the field and vice versa.

3. Can you explain the dipole moment mathematically?

The dipole moment is equal to the magnitude of either charge multiplied by the distance between the charges. It is also a vector quantity, with direction pointing from the negative to the positive charge.

4. What is the electric potential of a dipole?

The electric potential of a dipole is the sum of the potentials due to each individual charge. It is given by the equation V = q/(4πε0r)cosθ, where q is the magnitude of the charge, r is the distance from the dipole, and θ is the angle between the dipole moment and the line connecting the dipole to the point where the potential is being measured.

5. How is the electric field of a dipole calculated?

The electric field of a dipole can be calculated using the equation E = (1/4πε0)(2p/r^3)cosθ, where p is the dipole moment, r is the distance from the dipole, and θ is the angle between the dipole moment and the line connecting the dipole to the point where the field is being measured.

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