Experiment 1 - Magnetic Fields of Coils and Faraday's Law


Shown in the pictures below:

  • Stand with magnetic field sensor, linear motion rack, and rotary motion sensor
  • Spacer to mount magnetic field sensor on linear motion accessory
  • Helmholtz coils
  • Two search coils with parallel 10-kΩ resistors


Not shown in the pictures above:

  • Computer and interface
  • Voltage sensor
  • DC power supply to 2 A
  • Fluke multimeter
  • EZ angle protractor


This experiment consists of two parts. Part 1 involves checking the magnetic field produced by a current loop, while part 2 is an investigation of Faraday's Law. Most students cannot complete these two parts in one lab session, so you should choose which part you souls like them to perform. The default option (in which you do not express a preference) is part 2.


The most basic principle of electricity and magnetism is that charges exert forces on other charges. This picture is very simple if the charges are stationary: only the Coulomb force is present. Rather than describing the forces as action at a distance, however, we will use the field picture, whereby one charge creates a field, and other charges in the field feel forces from the field. If the charges are stationary, then only electric fields are involved; but if the charges are moving, magnetic fields also come into play. The field picture is used because the fundamental equations of electricity and magnetism — Maxwell's Equations — are much simpler when written in terms of fields than in terms of forces.

If a charge \(q\) is stationary, then it creates only an electric field \(\textbf{E}\):

\begin{eqnarray} \textbf{E} &=& q\textbf{r}/4\pi\varepsilon_0 r^3. \label{eqn_1} \end{eqnarray}

A charge moving at velocity \(v\) also creates a magnetic field \(\textbf{B}\):

\begin{eqnarray} \textbf{B} &=& \mu_0 q\textbf{v}\times\textbf{r}/4\pi r^3. \label{eqn_2} \end{eqnarray}

If a test charge is stationary, then it feels only an electric force \(\textbf{F}_E\):

\begin{eqnarray} \textbf{F}_E &=& q\textbf{E}. \label{eqn_3} \end{eqnarray}

If the test charge is moving, then it also feels a magnetic force \(\textbf{F}_B\):

\begin{eqnarray} \textbf{F}_B &=& q\textbf{v}\times\textbf{B}. \label{eqn_4} \end{eqnarray}

Often for the case of magnetic fields, the moving charges are part of an electric current. Consider a current \(i\) of charges \(q\) moving at an average drift velocity \(v_0\).

If the number of charges per unit volume in the material is \(n\), then the total charge \(\Delta q\) in the cylindrical volume of cross-sectional area \(A\) and length \(\Delta L = v_0\Delta t\) is \(nqA\Delta L\), since the volume of the element is \(A\Delta L\). Thus, the current \(i\) which passes through one cap of the cylindrical element is

\begin{eqnarray} i &=& \Delta q/\Delta t = nqA \Delta L / \Delta t = nqA(v_0\Delta t)/\Delta t = nqAv_0. \label{eqn_5} \end{eqnarray}

To find the magnetic field produced by this element of wire, we need to sum the \(qv_0\) terms in Eq. \eqref{eqn_2}:

\begin{eqnarray} \textbf{B} &=& \mu_0 \left(\sum q\textbf{v}_0\right) \times \textbf{r}/4\pi r^3. \end{eqnarray}

In the cylindrical element, the total number of charges is \(nA\Delta L\), so

\begin{eqnarray} \sum q\textbf{v}_0 &=& (nqA \Delta L) \textbf{v}_0 = i \Delta\textbf{L}, \end{eqnarray}

where we have used Eq. \eqref{eqn_5} in the last step, and have noted that \(\Delta\textbf{L}\) is in the same direction as \(\textbf{v}_0\). Thus, the magnetic field produced by this element of wire is

\begin{eqnarray} \Delta\textbf{B} &=& \mu_0 (i\Delta \textbf{L}) \times \textbf{r}/4\pi r^3. \label{eqn_6} \end{eqnarray}

Eq. \eqref{eqn_6} is the Law of Biot and Savart. (We call the magnetic field from this element \(\Delta\textbf{B}\) since there must be additional contributions to \(\textbf{B}\) from other parts of the wire carrying the current.)


Consider a circular wire loop of radius \(R\) and carrying a current \(i\). We are interested in the magnetic field at the center of the wire. (Of course, a real loop would need to be interrupted by a battery at some point to keep the current flowing through the resistance of the wire, unless the wire were a superconductor.)

For the current loop, the cross product \(\Delta\textbf{L} \times \textbf{r}\) is perpendicular to the plane of the loop, so all the \(\Delta\textbf{L}\) terms contribute to the magnetic field in the same direction. Note that in this case, the magnitude of \(\textbf{r}\) (the vector from the current element to the observation point) is equal to \(R\) (the radius of the loop). Therefore, to obtain the total magnetic field at the center of the loop, we need only to add all the \(\Delta\textbf{L}\) terms; no other quantities change as we move around the circle:

\begin{eqnarray} \int\textrm{d}L &=& 2\pi R. \end{eqnarray}

The magnitude of \(\textbf{B}\) for a loop is thus

\begin{eqnarray} B &=& \int \textrm{d}B = \mu_0 i \left(\int\textrm{d}L\right) R/4\pi R^3 = \mu_0 i (2\pi R) R/4\pi R^3 = (\mu_0/4\pi) (2\pi i/R), \label{eqn_7} \end{eqnarray}

and the direction of \(\textbf{B}\) is perpendicular to the plane of the loop. If you curl the fingers of your right hand in the direction that the current (which, by convention, is positive) flows, then your thumb will point in the direction of \(\textbf{B}\).

For a loop of \(N\) turns, the current is \(N\) times the current in one turn, so the magnitude of \(\textbf{B}\) would be

\begin{eqnarray} B &=& (\mu_0/4\pi) (2\pi Ni/R). \label{eqn_8} \end{eqnarray}

Now let us find the value of \(\textbf{B}\) at a distance \(z\) along the axis perpendicular to the plane of the loop.

Here the position vector \(\textbf{r}\) from the current element \(\textrm{d}\textbf{L}\) to the observation point a distance \(z\) along the axis is the diagonal vector whose magnitude is \((z^2+R^2)^{1/2}\). The magnetic field contribution \(\textrm{d}\textbf{B}\) is at an angle \(\theta\) with respect to the \(z\)-axis. As we integrate around the circle, only the vertical (\(\cos\theta\)) components add up; the horizontal (\(\sin\theta\)) components are canceled by an equal contribution on the opposite side. Also note that the angle \(\alpha\) is equal to \(\theta\), from the geometry theorem that if two lines meet at an angle, then two other lines, each perpendicular to one of the first two lines, make the same angle. Thus, \(\cos\theta = R/(z^2+R^2)^{1/2}\), and the contribution of the magnetic field along the \(z\)-axis is

\begin{eqnarray} \textrm{d}B_z &=& \textrm{d}B\,\cos\theta = (\mu_0/4\pi) (Ni\textrm{d}L/r^2) \cos\theta = (\mu_0/4\pi) [Ni\textrm{d}L/(z^2+R^2)] [R/(z^2+R^2)^{1/2}]. \end{eqnarray}

Again, when we integrate around the loop, none of the other quantities change, so using \(\int\textrm{d}L = 2\pi R\), we obtain

\begin{eqnarray} B &=& (\mu_0/4\pi) [2\pi NiR^2/(z^2+R^2)^{3/2}]. \label{eqn_9} \end{eqnarray}

You should be able to verify quickly that Eq. \eqref{eqn_9} reduces to Eq. \eqref{eqn_8} for \(B\) at \(z = 0\).


Conceptually, Faraday's Law tells us that changing magnetic fields induce electric fields. Mathematically, this law states that the emf \(\mathcal{E}\) — the integral of the electric field around a closed path — is equal to the change in magnetic flux \(\Phi\) through the path:

\begin{eqnarray} \mathcal{E} &=& \int \textbf{E}\cdot\textrm{d}\textbf{l} = -\textrm{d}\Phi/\textrm{d}t, \label{eqn_10} \end{eqnarray}


\begin{eqnarray} \Phi &=& \int \textbf{B}\cdot\textrm{d}\textbf{A} \end{eqnarray}

The minus sign in Eq. \eqref{eqn_10} reminds us of Lenz's Law: the emf is induced in such a direction as to oppose the change in magnetic flux that produced it.

In this experiment, we will be testing Faraday's Law by monitoring the emf induced in a small search coil of \(N\) turns, positioned in a changing magnetic field. For such a coil, the emf will be \(N\) times larger than the emf induced in one turn:

\begin{eqnarray} \mathcal{E} &=& -N\,\textrm{d}\Phi/\textrm{d}t. \end{eqnarray}

Furthermore, if the search coil is small enough so that \(\textbf{B}\) can be considered constant over the area, then

\begin{eqnarray} \Phi &=& \int \textbf{B}\cdot\textrm{d}\textbf{A} = \textbf{B}\cdot \int\textrm{d}\textbf{A} = \textbf{B}\cdot\textbf{A}. \end{eqnarray}

Combining these results, we obtain the version of Faraday's Law which will be tested in this experiment:

\begin{eqnarray} \mathcal{E} &=& -(\textrm{d}/\textrm{d}t) (\textbf{B} \cdot N\textbf{A}). \label{eqn_11} \end{eqnarray}

The emf \(\mathcal{E} = \int \textbf{E}\cdot\textrm{d}\textbf{l}\) is quite similar to the potential difference \(\Delta V = \int \textbf{E}\cdot\textrm{d}\textbf{l}\), and can be measured with a voltmeter or the voltage sensor of Data Studio.


  1. We will be using the Rotary Motion Sensor with the Linear Motion Accessory to map the axial field of a current coil. Arrange your apparatus as shown in the diagram below. We are not using the second coil until step 8. It may be present, but should not be hooked to a power supply yet.

  2. Double-click on the Rotary Motion Sensor in Data Studio, and insert its plugs into the appropriate digital channels. Also call up the Magnetic Field Sensor in the setup window, and insert the physical plug into analog channel A.

  3. For magnetic field measurements, the coil draws power from the DC power supply, which is wired through the Fluke Multimeter to measure the current to the coil. Be sure to turn the coarse and fine voltage controls of the power supply to zero before switching on the power supply; otherwise, the initial current may be too large and blow the fuse in the Multimeter. To measure a current, the Multimeter must be in series with the power source. Wire one lead from the ground of the DC power supply to the “Common” plug of the Multimeter. Wire a second lead from the mA plug of the Multimeter to one of the coil plugs. The second coil plug is wired back to the other output plug of the power supply. Wire directly to the coil leads; do not use the built-in series 1.2-kΩ resistor in this part of the experiment. Set your Multimeter to read on the 2000-mA DC scale.

  4. Set the voltage control of the power supply to zero, turn on the power supply, and adjust the coarse and fine voltage controls slowly until you obtain a current of 1000 mA = 1A.

  5. Make a preliminary check of the magnetic field of the coil. Carefully position the end of the Magnetic Field Sensor at the center of the coil. (We are not making computer use of the rotational sensor yet.) Use the controls on the Magnetic Field Sensor (hardware) to set it to radial mode and 10× reading. Double-click on the Magnetic Field Sensor on the computer screen, and check that the 10× measurement appears in the “Data” column. Drag a digits window to the Magnetic Field Sensor, and drag the 10× data to it. Turn “OFF” the current to the coil from the power supply, and click “Start” to see the gauss reading of the sensor. Push the “Tare” button on the sensor to zero the magnetic field reading. It is a good idea always to zero the Magnetic Field Sensor with zero current before recording measurements with the current on. Even when zeroed, the reading may jump around a bit. Now turn the power supply back on, and record the magnetic field reading and the current. (If your magnetic field measurements have negative values and you don't like this, then reverse the leads to the power supply.)

    Magnetic field (gauss) =                                              

    Current (mA) =                                              

    Compare your measured magnetic field with the calculated field from \(B = (\mu_0/4\pi) (2\pi Ni/R)\) (Eq. \eqref{eqn_8}). Remember that \(\mu_0/4\pi\) = 10-7 Tm/A, and you can read off the value of \(N\) (the number of turns) and \(R\) (the radius of the coil). Be sure to put \(R\) and \(i\) in the proper SI units. This formula gives \(B\) in teslas; convert to gauss using 1 tesla = 104 gauss.

    Calculated magnetic field (gauss) =                                              

    Percentage error =                                              

  6. Now arrange the stand with the Magnetic Field Sensor to map the magnetic field of the coil along its axis. Your arrangement should be such that as you turn the Rotary Motion Sensor, the Magnetic Field Sensor starts on one side of the coil, passes through the center of the coil, and moves beyond the other side — always staying on the axis of the coil. Double-click on the Rotary Sensor icon, and check the position measurement so it shows up in the “Data” column. On the computer screen, drag a graph to the Rotary Motion Sensor. Drag the position data to the \(x\)-axis of the graph, and the magnetic field data to the \(y\)-axis. Make a trial run by turning off the current, zeroing the magnetic field reading with the “Tare” button, turning on the current, clicking “Start”, and moving the sensor through the field by rotating the pulley wheel on the Rotary Motion Sensor. Rotate smoothly — although the speed is not important, since we are not plotting anything as a function of time. You should see a nice graph of the magnetic field plotted against the axial distance.

  7. After any readjustments, when you have a nice plot on the computer, you may print it out. DO NOT DISCARD THE DATA. Save it in a file, which you can retrieve, on the desktop. You may use it later for additional credit.

  8. Now wire the second coil in parallel with the first. You want the currents in the coils to flow in the same direction, so the magnetic fields of the two coils add in the space between the coils. Again, adjust the voltage of the power supply so the current to each coil is 1 A (with a total current of 2 A). Arrange the stand with the Magnetic Field Sensor to move along the axis of both coils, particularly covering the area between them. Take three measurements of \(B\) versus axial distance with the coil separations equal to \(0.5 R\), \(1.0 R\), and \(1.5 R\), where \(R\) is the radius of the coils. (In order to make the three graphs comparable, start each measurement with the end of the Magnetic Field Sensor at the center of one coil. If the leads of the second coil prevent you from moving it to the \(0.5 R\) distance from the first coil, how can you manipulate the apparatus to make this measurement possible?) Arrange a graph on the computer so that all three plots appear on the same graph aligned vertically. You may print this page out for your records.

    What coil separation produces the most uniform magnetic field between the coils?


    This arrangement is called Helmholtz coils, and is a method of producing a relatively constant, controllable magnetic field over a considerable volume of space. Of course, we have just measured the magnetic field along the axis, but the field is fairly uniform throughout much of the volume of space between the coils. We used this property of Helmholtz coils in the previous \(e/m\) experiment.


  1. Disconnect and set aside the power supply, the multimeter, and the stand with the Magnetic Field Sensor. Start with a new Data Studio display, being sure to save the data you might need for the additional credit.

  2. Wire the signal generator output of the Science Workshop interface to one of the coils, including the 1.2-kΩ resistor in series. We will call this the “field coil”. Since the resistance of the coil itself is negligible compared to the 1.2 kΩ of the resistor, the voltage of the signal generator divided by 1.2 kΩ gives the current through the field coil.

  3. Wire a voltage sensor (just an analog plug with two leads) to the 2000-turn search coil, and set it up in Data Studio. The search coils have 10-kΩ resistors in parallel to damp out oscillations. On the computer screen, double-click “Signal Output” to get the output voltage into the data column and bring up the signal generator control window. Set up a scope display with two traces to measure both the voltage output of the signal generator and the output of the voltage sensor. Select the signal generator voltage so the scope triggers on this signal.

  4. Set the signal generator to a 5-V triangle wave at 2000 Hz, turn it on, click “Start”, and make any adjustments to the scope so that you see a clear, stable triangle wave on the scope. Check the trigger setting if the wave on the scope is not stable. Adjust the trace speed by clicking the arrows on the \(x\)-axis of the scope, and the amplitude of the trace by clicking the appropriate arrows on the \(y\)-axis. Now when you bring up the search coil inside the field coil and in the same plane, and adjust the voltage sensor signal on the scope to greater sensitivity, you should see a square wave trace, possibly with some fluctuations. Notice that when you click “Stop”, the last scope trace remains fixed in the scope window, facilitating some of the measurements below.

  5. Faraday's Law (Eq. \eqref{eqn_11}) states that \(\mathcal{E} = -(\textrm{d}/\textrm{d}t) (\textbf{B} \cdot N\textbf{A})\), where \(\mathcal{E}\) is the emf induced in the search coil (which you are measuring with the voltage sensor). \(\textbf{B}\) is the magnetic field of the field coil, which is proportional to the current flowing through it; this current, in turn, is proportional to the triangle wave voltage from the signal generator. \(\textbf{A}\) is the area vector of the search coil, and \(N\) is the number of turns of the search coil. The cosine of the dot product between \(\textbf{B}\) and \(\textbf{A}\) is 1.0 if the search coil is in the same plane as the field coil.

    Study your two scope traces, and explain clearly, briefly, and neatly the shape and the phase of the voltage sensor output using the equation above.





PROCEDURE — DEPENDENCE OF \(\mathcal{E}\) ON \(\textrm{d}i/\textrm{d}t\)

  1. As you can see from Faraday's Law, \(\mathcal{E}\) is proportional to \(\textrm{d}i/\textrm{d}t\) (where \(i\) is the current through the field coil), since \(B\) of the field coil is proportional to \(i\). In turn, \(\textrm{d}i/\textrm{d}t\) is related to the frequency for a triangular wave. Write down the correct equation relating \(\textrm{d}i/\textrm{d}t\) to the frequency \(f\) and amplitude \(i_{\textrm{m}}\) for a triangular wave.


  2. Check this relation. Measure \(\mathcal{E}\) (by putting the smart tool on the square wave after you click “Stop”) at 500, 1000, 1500, 2000, and 2500 Hz. Enter your data in a new Excel worksheet, chart \(\mathcal{E}\) as a function of \(f\) with a clear title and labels on the axes. You may print out the graph for your records. Be sure you position the search coil in the same place at the center of the field coil when making each measurement.


  1. Devise a way to use the EZ angle protractor to adjust the dot product angle in \(\textbf{B} \cdot N\textbf{A}\) between the area vector of the search coil and the magnetic field (perpendicular to the field coil plane). The search coil should be at the center of the field coil for each measurement. It may help to have both lab partners holding parts of the equipment. Remember that when you click “Stop”, the signal will remain on the scope for easy measurement with the smart tool.

  2. Keeping the frequency and amplitude of the triangle wave constant, measure \(\mathcal{E}\) for dot product angles of 0°, 30°, 45°, 60°, and 90°. Record your data in a new Excel worksheet, compute the cosines with a fill-down operation, and chart \(\mathcal{E}\) as a function of the cosine of the angle. Title the chart and label the axes clearly. You may print it out for your records.


  1. We have been using the 2000-turn search coil up to now. Measure \(\mathcal{E}\) with a triangle wave of amplitude 5 V and frequency 2000 Hz, using the 2000-turn search coil at the center of the field coil and in the same plane. Now, what should \(\mathcal{E}\) be with a 400-turn search coil? Repeat the measurement with the 400-turn coil, and record your result below.

    Predicted \(\mathcal{E}\) with 400 turns =                                              

    Measured \(\mathcal{E}\) with 400 turns =                                              


  1. If the current to the field coil is a sine wave, what form of wave (including phase information) should the induced \(\mathcal{E}\) take? Verify this with your equipment.


ADDITIONAL CREDIT (5 mills maximum)

Before you start the additional credit, check that you have the following graphs with good data, titles, and labeled axes:

  1. Magnetic field versus axial distance for a single coil.

  2. Three magnetic field measurements versus axial distance for three different coil separations plotted on the same graph.

  3. An Excel chart of \(\mathcal{E}\) versus \(f\).

  4. An Excel chart of \(\mathcal{E}\) versus the cosine of the angle of the dot product.

The magnetic field on the axis at a distance \(z\) from a current coil of radius \(R\), \(N\) turns, and current \(i\) is given by Eq. \eqref{eqn_9}:

\begin{eqnarray} B &=& (\mu_0/4\pi) \left[ 2\pi NiR^2/(z^2+R^2)^{3/2} \right]. \end{eqnarray}

Go back to the data that you saved earlier for the magnetic field along the axis of the coil. Set up a calculation in Data Studio for the equation above, and plot it on the same graph as the magnetic field data so you can compare the theoretical magnetic field with the measured field. The reward is 5 mills if you have a neatly labeled printout, done without significant assistance from your TA. If you need assistance, you can “buy” varying amounts by giving up some of the 5 mills. Here are a couple of hints: \(B\) in the equation above is in teslas, but the measured field is in gauss, so include the conversion to gauss in your calculation of the theoretical field. You will also need to change the origin of the \(z\)-axis in the calculation to match the position of the center of the coil for your measured field.