Model 1.


Assume a small magnet moves along a line perpendicular to the plane of the loop of radius





-
Magnetic flux through the loop is the same when the magnet is at the points that are symmetric with respect to the plane of the loop:
. Therefore
is an even function.
-
Magnetic flux through the loop is maximum when the magnet crosses the plane of the loop:
.
- Magnetic flux vanishes at large distance from the loop:
as
.
Figure M.3 depicts a function that has these properties.
The flux of the magnetic field of the dipole moving along the axis of symmetry of the loop is given by (I.11) which we obtained by integrating the vector potential over the loop:

|
(M.1) |
It is easy to see that (M.1) satisfies all aforementioned properties and is of the shape shown in Figure M.3.
Our goal is to find the maximum value of the flux when the magnet passes through the center of the loop. As the magnet moves along the line perpendicular to the plane of the loop the induced emf in the loop as a function of time is (we do not watch for the signs here):
|
(M.2) |
where is the speed of the magnet. The derivative of the function of the shape of Figure M.3 is shown in Figure M.4.
has maximum at
its derivative
at
.If the sped of the magnet stays approximately constant as it moves through the vicinity of the loop
the shape of the
function mimics that of
as follows from Faraday’s Law (M.2) (see figure M.5). Let us denote the time when the magnet crosses the plane of the loop
. From (M.2) it follows that
Integrating emf (M.2) with respect to time from
to
yields
:
|
(M.3) |
Therefore the flux through the circular loop of the magnetic field of the small magnet in the center of the loop is the integral of the induced emf in the loop with respect to time as the magnet moves from a very distant point to the center of the loop along the loop’s axis – this is the area under the curve from negative infinity up to the time
where
crosses zero (
shown in green in Figure M.5). We may also integrate
from
to
:
|
(M.4) |
The values for obtained from (M.3) and (M.4) should, in theory, be the same. Unfortunately, iOLab device has small bias for such measurements on High Gain terminal. This means that the
curve is shifted a little – ether up or down. Because of this shift the values for
obtained through integration (M.3) from
to
(
in Figure M.5) and through integration (M.4) from
to
(
in Figure M.5) are slightly different. To exclude this bias, we will take the average of these flux values:
|
(M.5) |
For the magnet moving along loop’s axis of symmetry equating obtained through (M.5) with the value for
from (M.1) allows to evaluate the magnetic dipole moment
. You may compare it with the values you obtained using other techniques in Lab 5. I believe that the value for
obtained in this Experiment is more accurate (closer to the true value) than the values you calculated in Lab 5.
Model 2.



Our goal is to express the magnetic dipole moment in terms of emf amplitude
.
-
At first, we need to find
values where
is maximum (
and
in Figure M.4). From symmetry it is clear that
.
-
Differentiate (M.1) with respect to
to find the expression for
.
-
Differentiate
with respect to
to find
.
-
Solve equation
to find
and
where
attains its maximum/minimum values. Express
and
in terms of loop radius
.
-
Differentiate (M.1) with respect to
- Calculate
. It is clear that
.
-
From Faraday’s Law (M.2), it follows that
(M.6)
We will assume the magnet moves with constant speed over the interval
(note: the time interval
is very small). Therefore, we may approximate the speed of the magnet as
. Express
in terms of loop radius
and plug the speed of the magnet
and your expression for
in (M.6).
- Express magnetic dipole moment
from the resulting expression – you should have
expressed in terms of
,
,
, and
.
-
To exclude small iOLab’s bias use (
in place of
when calculating
from your experimental data (see Figure M.6)