Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

Force due to magnet

1 vote
426 views

The mid points of two small magnetic dipoles of length d in end-on positions, are separated
by a distance x, (x >> d). The force between them is proportional to $1/x^n$ . Then what is the value of n .

Magnetic fied (B) due to a bar magnet is propotional to $1/x^3$ so force should also proportional to this .

But the answer is n=4

asked Mar 20, 2017 in Physics Problems by koolman (4,286 points)
edited Mar 20, 2017 by koolman

2 Answers

0 votes
 
Best answer

This trick question trying to misguide you . The question says two magnetic dipoles at there end on set up behaving as a quadrapole.

Now a Quadrapole has its field as $B \ proportional$ to $1/x^4 $

similarly its potential also is proportional to $1/x^{n-1} = 1/x^3$

For more Information about a quadrapole refer Hyperphysics .
It has a good explanation along with a diagram .

answered Mar 20, 2017 by physicsapproval (2,320 points)
selected Mar 20, 2017 by koolman
sorry , I could not understand why it is quadrapole and why it is proportional to 1/x$^4$
2 votes

The magnetic field from a magnetic dipole is the same as from a current loop, and has the same form as the electric field from an electric dipole. The magnetic field along the axis of the magnetic dipole is proportional to $1/x^3$ [1].

In the case of the electric dipole, the fields from each charge are proportional to $1/x^2$ and cancel to 1st order of approximation if they are close together, leaving the 2nd order effect which is proportional to $1/x^3$.

If the magnetic field is uniform at the 2nd loop, there is no force on it, only a torque. When the magnetic field varies with distance there is a force (in addition to the torque) on the 2nd loop, equal to $F=\mu .\frac{\partial B}{\partial x}$ where the dot indicates scalar product [2]. This force either pulls the loops (dipoles) together or pushes them apart. The magnetic field from loop 1 varies as $B(x) \propto 1/x^3$ so the force of attraction (or repulsion) on loop 2 is also proportional to $1/x^4$.

Likewise, the force of attraction/repulsion between 2 electric dipoles is proportional to $1/x^4$.

[1] http://www.physicsinsights.org/dipole_field_1.html
[2] http://www.physicsinsights.org/force_on_dipole_1.html

answered Mar 20, 2017 by sammy gerbil (28,806 points)
edited Mar 20, 2017 by sammy gerbil
I can't understand how we get the $ F= \mu \delta B/ \delta x $
I also could not understand that .
Have you studied reference #2, the section entitled *Force on the Loop - z component of the force*? Or is that derivation what you don't understand?

It is easiest to see with the electric dipole, if you accept the analogy. With an electric field $E(z)=k/z^3$ from dipole #1, the forces on the 2 charges in dipole #2 are  $F_1=q_1E_1=-kq/(z-a)^3$ and $F_2=q_2E_2=+kq/(z+a)^3$, where + indicates repulsive and the -ve charge is nearer to dipole #1, which has the +ve charge leading. Here $2a$ is the distance between charges $\pm q$ in the dipole,

This calculation is similar to finding the field from the dipole. Add the forces, use the binomial expansion for the denominators, collect terms, and neglect higher order terms. You get a force proportional to dipole moment $p=2qa$ and also to $1/z^4$. This is the same as $F(z)=p\frac{dE}{dz}$.

The magnetic equivalent is a net force proportional to magnetic dipole moment $\mu$ and $\frac{dB}{dz}$.
I am sorry I mistook $ \mu $ as permeability , its magnetic dipole  moment. And the we used  F = dU / dx  as thats a " permanent dipole "  as in reference #2 . Am I thinking right then ?
Yes got that  :)
...