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6.1: Background Material

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  • Dipole-Dipole Force

    We already know that magnetic fields are not found in nature to exhibit a monopole (point charge) field (not yet, anyway, though we have been looking for over 100 years).  We also know that the north side of a magnet is attracted to the south side of another magnet, which means that attractive (and repulsive) forces exist between dipoles. We know that these forces between monopoles weaken with separation according to an inverse-square law. Our goal in this lab is to determine the power law that applies to the dipole-dipole force.  We will develop the theory here to form an hypothesis, and then test it with the lab activities.

    The simplest way to derive a power law for a dipole-dipole interaction is to use electric dipoles, set up with a simple geometry (in this case, aligning the dipole moments).  As usual with dipoles, we will only consider the far field – the case when the dipoles are separated by a distance much greater than the dipole separation of the two point charges.

    Figure 6.1.1 – Two Dipoles


    Treat the center of the left dipole as the origin, so that the separation of the dipoles is \(r\), and by assumption, \(r\gg a\).  The force on the right dipole is a sum of four forces from the left dipole (one Coulomb force on each point charge by each of the other two point charges).  It may be surprising at first that the net force doesn't come out to be equal to zero, and once this becomes clear, the "hard part" is to do the algebra to get a common denominator for the expression for the force and then to impose the condition that \(r\gg a\) to get a single power of \(r\) to describe this force.

    Log-Log Graphs

    When we perform the experiment, our data will result in a force-vs-separation curve that (of course) shows that the force gets stronger as two bar magnets (dipoles) get closer together.  But it can be a bit tricky to determine a power law from a graph.  In a previous lab, we had two theoretical functions to check, so we simply "linearized" both and checked to see which gave a better best-fit line. In this lab, we will remain agnostic about the power law (other than assuming a simple power law exists), and use a common technique for determining the power.

    We assume that data satisfies a function that has this form:

    \[f\left(r\right) = a r^\alpha\]

    We can, as we did previously, test many different values of \(\alpha\) by graphing the data of \(f\) vs. many different powers of \(r\), and the one that gives a straight line tells us the value of \(alpha\).  But suppose we take the logarithm of both sides of this equation (any base at all can be used for this logarithm, but we will use the natural log):

    \[\ln f = \ln \left[a r^\alpha\right] = \ln a + \alpha\ln r \]

    This shows that if we graph the logarithm of the dependent variable against the logarithm of the independent variable, then we will get a straight line, with a slope equal to the power \(\alpha\) of the functional relation (as a bonus, we also get the multiplicative factor \(a\) as a vertical axis intercept).

    Deflection of a Moving Charge

    In the second part of this lab, we will be looking at the deflection of electrons passing through a magnetic field.  The source of electrons is a cathode ray tube (CRT), and it is passing between two Helmholtz coils (coiled loops of wire that carries an electric current).  A magnetic field is produced by the current in these coils, and for the region through which the electrons in the CRT are traveling, this field is approximately uniform and perpendicular to the direction of motion of the electrons.  We can then vary (and measure) the current through the coils to vary the magnetic field, and observe the deflection of the electron beam (which makes a green dot on the front screen that we can observe).

    Figure 6.1.2 – CRT & Helmholtz Coil Setup

    Screen Shot 2020-11-06 at 1.29.26 PM.png

    There is a lot of math needed to link our two direct measurements of the electric current through the coils and the deflection distance of the beam.  Let's tackle that here...

    The force on the electron moving perpendicular to the field is given by the usual Lorentz force:

    \[\vec F = q \vec v \times \vec B \;\;\; \Rightarrow \;\;\; F=qvB \]

    The quantity \(q=e\) is the charge of the electron. We have perhaps not yet covered the strength of a magnetic field near the axis of two Helmholtz coils (whose fields are pointing the same direction) yet. We will use this in the next lab as well, but for the purposes of this lab, we'll just write it down:

    \[B\left(I\right) = \left(\dfrac{\mu_oNr^2}{\left(r^2+z^2\right)^{\frac{3}{2}}}\right)I\]

    In this equation, \(I\) is the current through the coils, \(r\) is the radius of the coils, \(N\) is the number of turns of wire in each coil, \(z\) is the distance along the axis from the plane of a coil, and \(\mu_o\) is a constant called the permeability of free space:

    \[\mu_o = 4\pi \times 10^{-7} \frac{T\cdot m}{A}\]

    With the electron beam initially moving horizontally, and the deflection due to gravity negligibile, the amount of deflection of the beam due to the constant magnetic force as a function of time is:

    \[\Delta y = \frac{1}{2} at^2 = \frac{1}{2} \left(\frac{F}{m}\right)t^2 \]

    The electrons are accelerated from rest horizontally inside the CRT before reaching the magnetic field, by dropping them through a voltage we'll call \(V\).  Conservation of energy gives us the speed at which the electrons emerge, which (as we said above) is horizontal, so:

    \[\frac{1}{2}mv_x^2 = eV\;\;\; \Rightarrow \;\;\; v_x = \sqrt {\frac{2eV}{m}}\]

    The quantity \(m=m_e\) is of course the mass of the electron.  We also have the dimensions of the CRT, so we know how far they travel to the screen \(L\) through the magnetic field, giving us the time of travel:

    \[t = \frac{L}{v_x} = L\sqrt {\frac{m_e}{2eV}}\]

    Putting several of these equations together finally gives us a formula that relates the current and the deflection:

    \[\Delta y  = L^2\sqrt {\frac{e}{8m_eV}}\left(\dfrac{\mu_oNr^2}{\left(r^2+z^2\right)^{\frac{3}{2}}}\right)I\]

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