AMPERE’S LAW

APPARATUS DC power supply, Hall Effect magnetic probe, Fluke high resolution voltmeter, solenoid, computer, digital ammeter, meter stick

Magnetic Monopoles fit perfectly into electromagnetic theory as sources of B fields. (Recall that B is a cross-product vector, with direction given by the right hand rule by convention. This contrasts with the electric field E, a polar vector.) A magnetic monopole would be an isolated north or south magnetic pole with a radial magnetic field pattern, directed out or in, respectively, like the E field of a positron or an electron (point-source electric monopoles). However, despite intense effort, no magnetic monopole has ever been observed - a challenge for theorists and cosmologists.

This leaves magnetic dipoles, a north-south pole combination, as the fundamental observed source of magnetism. The corresponding field pattern can be revealed by iron filings aligned in the field of a bar magnet. Cutting a bar magnet does not produce separate north and south poles but rather, two bar magnets. Electric charges that are in motion, macroscopic (transformer currents etc.) or microscopic (aligned orbiting or spinning atomic electrons in a permanent magnet) are the observed sources of B fields. Current in a coil produces a field pattern of dipole shape, with field lines leaving one end of the “tunnel”, looping around externally, and reentering the “tunnel” at the other end. These field lines are continuous – there are no magnetic monopoles for them to begin or end on.

For a long straight wire carrying current I the B field line direction is tangent to centered circles and the B magnitude is  = ₀I/(2R). The circle’s length is 2R, so B_circle x 2R = ₀I. This easy result is only a special example of Ampere’s law, which states that the sum around any closed path of tangential B field component times path line element (“line integral”) equals ₀I, where I is the net current passing through the loop:

in general. Equation 1

(I must be understood as NI_ammeter if the current passes N times through the path.)

If there is no net current within the closed path, the closed integral is zero. This does not necessarily mean there is no B field present along the line integral, or no currents enclosed. Rather it means that the dot product with the field direction sums to zero. (“Up” and “down” currents through the enclosed surface must be assigned opposite signs.)

Think of two adjacent wires with equal and opposite currents. The closed line integral surrounding them is zero. If the closed line integral is not zero, you know that there is a net current within the closed path which is generating a magnetic field.

SOLENOIDS

An application of Ampere’s Law involves a solenoid (a wire coil wound on a cylinder) with:

N = number of turns of solenoid (dimensionless)

R = radius of coil (meters)

I_ammeter = current through solenoid (amperes)

L = length of solenoid (meters).

It can be shown that the B field intensity at the center of the solenoid is

Equation 2 and at the end of the solenoid is

Equation 3.

N can be determined from B measurement at either of these points, as well as from the line integral of tangential B along any loop passing through the coil.

HALL EFFECT A magnetic field can be measured with a Hall Effect probe. In the diagram below, a current, I, is transmitted through a silicon semiconductor. The potential between the top and bottom points is zero until a perpendicular magnetic field is applied which exerts a force on the moving charges. If the current consists of positive c

Figure 1: Hall Effect

You will use a Hall Effect magnetic field detector to measure the magnetic field. The detector is mounted at one end of a clear plastic block that can be oriented in a magnetic field. The output is amplified and recorded by a sensitive voltmeter. Evaluation of the line integral of the B-field’s parallel components for two or more closed paths through a solenoid will determine the number of solenoid turns N, by application of Equation 1.

In evaluation of the line integral we will approximate infinitesimal line elements dl by finite elements Dl:

Do not reverse the direction of circulation around the loop during your summation.

Keep magnetic material (steel watch bands, bracelets, etc.) away from the experiment. Besides the solenoid, currents in power supplies, computers, etc. produce magnetic fields

Calibration and Zero-offset Procedure

Calibrate the Hall probe by using the Earth’s magnetic field. The probe will be rotated 360 degrees to determine the effective “zero” reading for the probe. (Use the inscribed line on the bottom of the clear plastic holder to align the probe.) The amplifier has an offset somewhere between 1 and 10 volts (the zero field reading). For example, as you rotate a probe horizontally through the Earth’s magnetic field, you will see a range of Hall voltages V_H, say, 2.409V and 2.391V. This probe, thus, has a zero offset, V_o of 2.400V. Positive readings of V_H - V_o indicate a positive field, while negative readings of V_H - V_o are negative fields. Then use the amplitude of the variation to calibrate the probe, V_H - V_o, to the units of B in Gauss. The full peak-to-peak amplitude will be set to 0.6 Gauss which will give you the conversion factor, b( Gauss/Volt): B(Gauss) = b(V- V_o). In this example b=0.6 Gauss/(2.409V-2.391V) = 33.33 Gauss/Volt. {1 Tesla (mks unit) = 10⁴ Gauss.}

Calibrate on the “compass” diagram that positions the probe in 12 directions, separated by 30 degrees. Before turning on the 5V supply for the Hall probe, disconnect the +12 V lead going to the solenoid. (A single power supply provides both solenoid and Hall probe voltages.) Turn on the supply and set the Fluke multimeter to (Volts DC). Allow a few minutes for electronic stabilization. Record and point plot V_Hall vs. angle (call position 12 zero degrees). Calculate and record this first zero offset, V_0,1 from the average V_H reading. Then calculate the conversion constant b from the range of voltage values.

Figure 2: Earth Field Component Normalization, and First Hall Probe Zero: V_0,1

B vs. I Proportionality Test.

Next, turn off the power supply and connect the solenoid red lead to the ammeter + input. Connect the ammeter – input to the + terminal of the 2 - 20V variable outlet of the current supply. Connect the solenoid black lead to the current supply – (ground) terminal. Turn the power supply to around 10 volts (wait a couple of minutes for stabilization). The ammeter should read a positive current. Place the Hall probe inside the solenoid “tunnel” in the path direction so that the probe is stable in position and orientation. Don’t let the cord drag it.

Vary the solenoid voltage SV from 0 to 12 volts and record (directly into a computer spreadsheet if available) data pairs +I (solenoid current) and V_Hall. Turn off the current supply and reverse its output leads. Record V_Hall data for – I (SV from 0 to 12 volts). Scatter Plot V_Hall vs. I (enter + or – signs) and fit a straight line. Reject any data points which are obviously off trend and repeat fit. Record V _0,2 the I = 0 value of V_Hall. Use V_0,2 rather than V_0,1 in later zero-offset corrections of Hall probe voltages.

(V_H–V_0,2) should be proportional to the parallel component of the magnetic field (B_ll) at the measured point (if you aim right), whereas V_H is merely linearly related. From the curve fit below, V_0,2 is 2.4742 volts for that particular Hall probe.

Figure 3 Hall Probe Linearity Check, and Second Hall Probe Zero: V_0,2

3. Closed Line Integral Enclosing Non-Zero Net Current

The closed path which passes through the solenoid tunnel closed path (loop “B”) will give a non-zero integral which contains information about the product of current I and the number of coil-turns N in the solenoid. A fixed current I will be measured with an ammeter, so determination of the closed path line integral will enable calculation of N, the number of coil turns. (We assume Ampere’s Law instead of testing it, lacking independent knowledge of N.)

The product of current I and coil turns N can also be related to the point field strengths B_E and B_C at the end and center of the coil (see Equations 2 and 3). And since both B_E and B_C and also the closed-loop line integral are proportional to NI, we can relate the experimental Ampere’s law line integral to the point magnitudes B_E (or B_C) and to the geometric parameters of the solenoid, R and L.

From Eq. 1, 2 and 3: , or = _.

(Remember: You don’t know the number of turns N, but can measure the loop integral.)

Thus, the line integral can be predicted theoretically from the coil geometry and one point measurement of the B-field, either B_E or B_C. To verify these relations among B_E, B_C and the Amperian loop integral you need to measure the coil radius R, length L,, and the point field magnitudes B_E, and B_C ,and perform the line integral of field components along any closed paths that encompasses the solenoid wires. Use an average value of coil radius R.. B_E and B_C point along the solenoid tunnel axis; aim the Hall probe accordingly.

Connect for positive current, with solenoid supply voltage around 10 volts. (Different groups use different I values.) Follow the arrow direction completely around loop B, placing the Hall probe (at end of plastic block) on successive 2 centimeter marks, aligned with the path. Don’t measure twice at the start position! Wait two or three minutes for stabilization, then record in a spreadsheet V_Hall vs. segment number. Subtract V_0,2 from all V_H values to obtain V_B-field. Multiply all V_B by the product of b (magnetic field conversion constant obtained previously from earth field) and line segment length (0.02 meters) to obtain Bdl differential values. Sum these (algebraically) to obtain the loop B*dl integral in gauss-meters, a non-mks unit. Record the fixed current I. Plot the individual B_lldl values vs. line segment number. (Lack of right-left symmetry results from corner start, which is not symmetrically located with respect to the solenoid tunnel.)

Measure the solenoid length L and mean coil radius R. Measure the solenoid end and center point Hall voltages. (Orient the probe along the solenoid axis. Notice that the component value becomes zero if the probe is rotated into a perpendicular position.) Calculate the B-field values: B_End and B_Center. Compare the ratio B_C/B_E to the solenoid geometric expression ratio from Equations 1 and 2. (Both B_C and B_E are proportional to NI, so the ratio of the two depends only on the solenoid (average) radius R and its length L. The ratio of B field values B_C/B_E will be just the ratio of the corresponding C and E values V_B = (V_Hall-V₀); it is not necessary to convert.)

From your field-component loop integral, current I_ammeter and m₀, calculate N, the number of coil turns. (See Equation 2.) (First, convert your line integral value in gauss-meters to mks units Tesla meters.)

Don’t confuse “loop B” with B, the magnetic field!)

Figure 5: Line Integral Differentials for Closed Path Through the Solenoid

In the previous figure are four groups of line integral contributions (groups 1 – 4, from left to right), from the four rectangular path sides (a, b, c, d).

Group 3 (8 line segments of 0.02 meters each) involves the contributions from the path through the solenoid tunnel on axis, where the B field is strong and parallel to the path. Group 1 involves the parallel side outside the tunnel in the external dipole field pattern; the B field is weaker.

Groups 2 and 4 (6 segments each) involve the two path sides which are perpendicular to the tunnel axis.

Note that all contributions are positive. Compare with loop C, Figure 6. (Reversing the direction of the path, or reversing the current (but not both) would invert the pattern.) “Wrapping around” the data, i.e. shifting the pattern by four segments (i.e., starting at segment 5) would yield a symmetric plot, but lose the association of contiguous segment groups with the path sides.

90 degree probe rotations occur at corners, between segments:

28-1 (side d -> side a, group 4 -> group 1),

8-9 (side a -> side b, group 1 -> group 2),

14-15 (side b -> side c, group 2 -> group 3), and

22-23 (side c -> side d, group 3 -> group 4).

The sudden jump between segments 14 and 15 and between 22 and 23 is due mainly to ninety degree probe reorientation, rather than to field strength change.

4. Line Integral Enclosing Canceling Currents

Figure 6: Line Integral Differentials for Closed Path Surrounding the Solenoid

There are six groupings in the previous figure: Groups 1 – 6, left to right. There are four sides to the rectangular path: a, b, c, d.

Groups 1 and 4 (9 segments each of 0.02 meters) represent contributions from the path sides (a and c) parallel to the tunnel axis. They have different magnitudes because side c is closer to the solenoid than side a. They have different signs because one side (a) is traversed generally parallel to the B-field, the other (c) anti-parallel.

Paired groups 2 and 3 together represent a single path side (b) of 12 segments, perpendicular to the tunnel axis. The field component direction reverses at the tunnel axis crossing between segments 16 and 17 (not 15 and 16, because side c is closer), as this side is traversed. Paired groups 5 and 6 together represent the other long side (d), with similar reversal between segments 35 and 36.

90 degree probe rotations occur at corners, between segments:

42-1 (side d -> side a, group 6 -> group 1),

9-10 (side a -> side b, group 1 -> group 2),

21-22 (side b -> side c, group 3 -> group 4, and

30-31 (side c -> side d, group 4 -> group 5).

Note the high degree of cancellation in the pattern algebraic sum.

For the same current as for loop B, take differential line integral data for the large, rectangular closed path (Loop C) that surrounds the solenoid without passing through it. (Note that loop C is not symmetrically placed with regard to the solenoid.) There is sometimes a strong field component as you move around the path. But when you do the summation the components (some are positive, others negative) should effectively cancel one another (within errors), as predicted by Ampere’s Law. Note that the Hall probe might indicate small field component for large field magnitude, e.g. on the solenoid axis with probe orientation perpendicular to the axis. Both loops B and C provide examples.

It is important to realize that the line integral of B_ll should sum to zero for any path surrounding the solenoid - a centered circle, an offset circle, an irregular path. Rectangular paths are used only for alignment convenience

Record V_H values directly into the computer, if available. Sum and analyze as before.

5. Closed Line Integral Enclosing No Currents.

Do the integral test for the small closed loop A that is to the side of the solenoid. The loop contains no currents, so what do you expect the result to be? Plot as above and interpret your graph.

6. Path Independence Triangular path through solenoid enclosing net current

If time permits, evaluate the closed line path integral for another self-selected path D, enclosing net current. Straight line segments afford the easiest probe alignment. A triangle with one side along the solenoid axis would be a different geometry. Use a meter stick as a guide, and to measure segments. Record and calculate as before. Print the column graph. Compare with Loop B.

For direct comparison of closed path integrals, you should use the previous current. If you don’t do this, correct one or the other loop integral by the I ratio. Remember the loop integrals are directly proportional to I.

Preliminary Questions Name/Course-section/Date

Look up the expressions for point solenoid magnetic field values B_E and B_C.

What would be the ratio B_E/B_C if the solenoid were very wide but thin like a pancake? (Assume R>>L in the formula.) Show each step in your math.

What would be the values if the solenoid were very long and skinny (L >> R).

Is there a value of R/L for which B_E = B_C? If so, what is the ratio?

Relative to the one-loop integral, what happens to the line integral of B_tangential ^. dl if you loop through the solenoid seven times (in the same direction)?

What happens to the line integral of B_tangential ^. dl if you loop through the solenoid seven times (in the same direction) and three times in the reverse direction?

What happens to the line integral of B_tangential ^. dl if you loop through the solenoid ten times (in the same direction), seven with forward current I and three with reverse current -I?