6/06 Ampere's Law. Ampere's Law. AndreMarie Ampere in France felt that if a current in a wire exerted a magnetic


 Tracey Manning
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1 About this lab: Ampere's Law AndreMarie Ampere in France felt that if a current in a wire exerted a magnetic force on a compass needle, two such wires also should interact magnetically. Beginning within a year of Oersted's discovery, in a series of ingenious experiments he showed that this interaction was simple and fundamental parallel (straight) currents attract, antiparallel currents repel. The force between two long straight parallel currents was inversely proportional to the distance between them and proportional to the intensity of the current flowing in each. (See below on Ampere.) Today, we posit a magnetic field force intermediary, which is produced by currents (macroscopic or microscopic) and which, in turn, exerts force on other currents. The experimental properties of magnetic fields differ in a fundamental way from those found for electric fields the simplest magnetic field pattern is dipole, whereas the simplest electric field pattern is monopole (i.e., radial). The magnetic dipole is irreducible cutting a bar magnet does not produce separate + and magnetic monopoles, but rather two dipoles. Ampere's result for a long, straight current carrying wire implies a corresponding magnetic field which is concentric around the wire, proportional to the generating current and falling off inversely with distance. It is easy to see that the line integral along any one of these circles is proportional to the current within (2p r) * (1/r) remains constant, the sign depending on the current sign. It follows, with a little more work, that the line integral is not changed by path distortion. And, since the magnetic fields produced by several currents are the (vector) sum of those produced separately, the (algebraic) total current within a closed loop governs the line integral, independent of path. Apparatus: Solenoid and path integral board, DC power supply, cables, Hall Effect magnetic sensor, multimeter (two), meter stick, Graphical Analysis program References: Cutnell & Johnson: Physics, 6 th Ch. 21 Serway & Beichner: Physics for Scientists and Engineers, 5 th v2 Ch. 30
2 Figure 1 Magnetic field lines from two long straight parallel wires, carrying equal currents in opposite directions. B dl = 0 around the rectangle because there is no net current through the rectangular loop. But, for a loop encircling only one of the wires, B dl 0 From: mples.html
3 Figure 2 Magnetic field lines from two long straight parallel wires, carrying equal currents in opposite directions. B dl 0 because there is net current through the rectangular loop.
4 Figure 3 Magnetic field lines from a coil of N = 4 circular turns. A closed loop that passes through the coil will have B dl = 0 x 4I Teslameters, where I is the coil current in amperes. From A closed loop not passing through the loop will have B dl = 0. Introduction Ampere's Law states that the line integral of B and dl over a closed path is 0 current enclosed in that loop: times the B dl = 0 I enclosed µ 0 = 4π x 107 Teslameters/ampere For a long straight wire, the magnetic field is given by: B long straight wire = 0 I 2 R
5 (For a long straight wire carrying current I the B field line direction is tangent to centered circles. Along those circles, B is constant, so we can pull it out of the integral. The remaining integral is simply the circumference of the circle, which is 2 R, s o B long straight wire x ( 2 R )= 0 I. In your experiment, I must be understood as NI multimeter since each single coil wire of the solenoid contributes I to the enclosed current: B dl = 0 N I = 0 N I multimeter in general. Equation 1 where the dot denotes parallel component. If there is no net current within the closed path, the closed line 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 differential B dl elements sum to zero. Note that 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 both 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. In this lab you could actually sum up the contributions of B dl over such a path around a solenoid, to check if their sum does indeed equal 0 times the current enclosed by your path, if we knew the number N of solenoid turns (which we don't). Instead, you will assume the correctness of Ampere's Law and calculate N. The magnetic field around the solenoid will be determined by a magnetic sensor (which needs magnetic field calibration and voltage offset determination). You will measure the output voltage of the sensor (which is proportional to B after offset subtraction) using one multimeter. The current I through the solenoid will be measured directly by another multimeter.
6 Figure 4 Apparatus board with solenoid of (unknown) turns N, integration paths and calibration compass rose, power supplies and meters Some solenoid properties 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 multimeter = current through solenoid (amperes) L = length of solenoid (meters). Point B field intensities are calculated to be: B C = 0 N I ammeter 4R 2 L 2 Equation 2 at the middle of the solenoid (on axis)
7 B E = 0 N I ammeter 2 R 2 L 2 Equation 3 at the end of the solenoid (on axis) N can thus be determined from these point values of B (since we can measure all other parameters), 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 sensor. 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 charged carriers, a positive charge will accumulate at the lower end of the conductor. Negative carrier flowing in the same direction as I will induce a negative charge at the lower end. Thus, the Hall Effect can distinguish the charge of the carrier! (In this figure, the convention that current is a flow of positive carriers is shown.) In any case, a small but measurable potential is induced by the magnetic field. If the field is reversed, so is the polarity of the induced voltage. Figure 5 Generation of a Hall voltage from a current flow in a magnetic field. Voltage sense depends on sign of charge carriers. 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 Bfield 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 l :
8 0 N I ammeter = B dl B l = B l cos l for every contribution is simply the distance between the line segment markings. The angle in the dot product is always 0 degrees, since we are always orienting the black line along the path, so the above equation reduces to: 0 N I ammeter = B l This assumes that the loop passes through the solenoid, not around it. 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 Procedure SAVE GA FILES FREQUENTLY FOR CRASH RECOVERY Preliminary 1: Hall probe zero offset determination The Hall probe operation does not allow it to read zero for zero solenoid current. We need the zero current Hall probe reading value, V 0. We can turn off the solenoid current, but we can't turn off the earth's field, so we will measure V Hall for several solenoid currents I (+ and ), making the earth field contribution negligible, then plot and extrapolate V Hall readings to zero current to obtain the offset V Connect the red cable into the variable side of the power supply and turn the black knob fully counterclockwise (current is a minimum); this will allow a current to flow through the solenoid. Place the Hall sensor holder inside the center of the solenoid, oriented along the its length (9 o'clock compass direction). See figure below.
9 Figure 6 Finding the Hall voltage offset correction 2. Turn on the Keithley multimeter and set it to ammeter mode. This will enable you to measure the current through the solenoid. Keeping the Hall sensor fixed and motionless, measure V solenoid current as a function of current I as you increase the current from 0.04 A to 0.18A (approximate values). Reverse the power supply (variable side) leads. (Don't turn solenoid current supply off Hall probe voltage should remain on to maintain stabilization.) Record from A to A. B Open the Graphical Analysis program Ampere.GA3. Enter your data on Page 5 and observe the plot on page 6. Do a linear curve fit: V (solenoid current) = ax+b, then V 0 = b. Equation 4 b is V 0, the zero current value of Hall probe voltage ( zero offset correction). Preliminary 2: Magnetic Field Calibration of Hall Probe Voltage Calibrate the Hall sensor by using the Earth s magnetic field so that we can convert the Hall voltage (VV 0 ) to a magnetic field (Tesla). Note that the Hall sensor measures the component of any magnetic field along the black line inscribed on the clear plastic sensor holder. It does not necessarily measure magnetic field If you aim it
10 perpendicular to a strong magnetic field, it will give zero (after V 0 offset correction.) Since we do not know the direction of the horizontal field of the earth, the sensor will be rotated 360 degrees to determine the effective zero reading for the sensor. The sensor will register the highest voltage (magnetic field) when it is oriented towards the Earth's North Magnetic Pole and the lowest voltage (magnetic field) when it is pointed towards the Earth's South Magnetic Pole, which should be 180 degrees from North. At angles in between them the voltage will register some intermediate value, but never zero. The earth's field is a vector so the shape of V compass vs. compass angle θ should be sinusoidal V compass = A*sin(B*θ+C)+D The amplitude A (volts) of the compass curve sine fit yields a proportionality constant k in terms of the assumed horizontal component of the earth's field: k = B horizontal (Tesla)/Amplitude (volts), where B horizontal = 0.3x104 Tesla, in Piscataway. k = 0.3x104 /Amplitude Equation 5 The offset value D of the sinusoidal fit also determones V 0. We won't use this (the value you obtained previously is more accurate), but you could compare with the previous V 0 value. Enter V compass vs. angle (degrees) in the GA data columns on Page 5 and observe the graph on Page 6. Fit the sine and determine k from Equation 5. Enter your k and V 0 values in the three GA B calculated column definitions on Page 5 (for loops A, B and C), substituting for the placeholder values. Finally, after you enter your values, Graphical Analysis will calculate tangential B field components as B tangential = k (V Hall V 0 ). (This assumes that you orient the Hall probe correctly.) Field calibration in more detail: 1. Make sure the red cable is disconnected from the variable side of the power supply so that the solenoid has no current going through it. The regulated side of the power supply (5 V) powers the Hall sensor, which should always be on during the experiment
11 in order to maintain stability. Turn on the multimeter and set it to measure DC voltage  This will read the Hall sensor voltage which is related to magnetic field. Hall voltage readings should be zero before the power supply is turned on, since the power supply not only supplies the solenoid current; it also powers the Hall sensor. 2. Turn on the power supply. The Hall voltmeter should now register a voltage around 2.4 V. Place the Hall sensor, mounted on the clear plastic holder, on the ZERO FIELD COMPASS flat on the path integral board with the black cable. Rotate the holder and watch the voltage vary  which direction is Earth Magnetic North (golf course, ARC building, Physics Lecture Hall, Davidson)? 3. Take voltage data V compass as you rotate the sensor holder through 360 degrees, from 12 o'clock to 11 o'clock. See figures below. Fit a sine; calculate and record field calibration factor k as specified above (Equation 5). (The fit offset should be approximately your previously determined V 0 obtained by varying the solenoid current.) Figure 7 Mapping the earth's horizontal field component. C. Closed line integral enclosing nonzero net current (Loop B) Do this first.
12 Figure 8 Measuring a tangential magnetic field component Check the default dl values for each loop in the appropriate GA data columns on Page 5. If you plan on using different segment lengths, reset. Default values are set as 0.02 meters for loops B and C, and 0.01 meters for loop A, as marked on the solenoid board. Set the power supply so that it puts out ~ 0.15 A to the solenoid coil. Maintain this solenoid current for the remainder of the lab. Place the Hall sensor holder on Loop B, putting the tip (the sensor itself) right at the labeled starting point. Follow the path along the arrows indicated, taking voltage readings at each 1 or 2centimeter line segment marking. Make sure the black line on the sensor holder is oriented along the path at every point. Enter your voltage data vs. Segment # in Ampere.GA3 ( Vb raw data ). Remember not to measure the start position twice. Examine the bar graph for Loop B. Use the AnalyzeIntegral function to sum up the B.dl contributions ( B dl B l ) and record. From Equation 1, calculate and record the number of turns N Loop B in the coil.
13 Figure 9 Entering parameters in a Graphical Analysis calculated column definition for LoopA Bfield tangential component. Remove and replace placeholder values of V 0 and k with your own values. D. Closed line integral enclosing zero net current (Loop C) Repeat above for Loop C (zero net current enclosed). Use same solenoid current. E. Comparison of end and center solenoid field strengths 1. Measure the length and radius (average) of the solenoid using the meter stick. Use mks units. 2. Measure axial values of the magnetic field, with the sensor holder oriented along the solenoid axis (lengthwise), at the center and at the end of the solenoid (these are clearly marked on the part of the path inside the solenoid). Record. Keep in mind that the sensor is at the tip of the sensor holder. 3. From each measurement, calculate N: N center and N end. Compare with N Loop B. F. Closed line integral enclosing zero net current (Loop A)
14 If time permits, repeat again, this time for Loop A (zero net current enclosed). Use same solenoid current as previously. Report You have all the data to plot and compare the line integral differentials for Loops A, B and C. You may view all of your line integral loop data simultaneously on Page 4. A bar graph would be confusing ; it is deselected and point protectors substituted. Print a Page 1 bar graph for Loop B. Annotate it to explain relative signs and changes in terms of the path followed and of changes in Hall probe orientation. (Print additional graphs as directed.) On it, give your calibration parameter values of k (Tesla/volt) and V 0 (volts), your coil current I (amperes), your line integrals for loops A, B, C (Teslameters), line integral ratios C:B and A:B, your three values for N (solenoid turns): N Loop B, N center, N end. Give point B values (Tesla): B center and B end. Show any calculations on back. Look at the coil and estimate the # N of windings (estimate (number of layers) x (turns per layer). Is agreement w/ampere's Law reasonable? Record and discuss. Ampere and Electromagnetism Hans Christian Oersted was a professor of science at Copenhagen University. In 1820 he arranged in his home a science demonstration to friends and students. He planned to demonstrate the heating of a wire by an electric current, and also to carry out demonstrations of magnetism, for which he provided a compass needle mounted on a wooden stand. While performing his electric demonstration, Oersted noted to his surprise that every time the electric current was switched on, the compass needle moved. He kept quiet and finished the demonstrations, but in the months that followed worked hard trying to make sense out of the new phenomenon. But he couldn't! The needle was neither attracted to the wire nor repelled from it. Instead, it tended to stand at right angles.. In the end he published his findings (in Latin!) without any explanation.
15 AndreMarie Ampere in France felt that if a current in a wire exerted a magnetic force on a compass needle, two such wires also should interact magnetically. Beginning within a year of Oersted's discovery, in a series of ingenious experiments he showed that this interaction was simple and fundamental parallel (straight) currents attract, antiparallel currents repel. The force between two long straight parallel currents was inversely proportional to the distance between them and proportional to the intensity of the current flowing in each. {Only for those pursuing the math: this is not the basic force formula. Given two short parallel currents I 1 and I 2, flowing in wire segements of length L 1 and L 1 and separated by a distance R, the basic formula gives the force between them as proportional to I 1 I 2 L 1 L 2 /R 2 ( It gets further complicated if the currents flow in directions inclined to each other by some angle). To find then the force between wires of complicated shape that carry electrical currents, all these little bitty contributions to the force must be added up, i.e. integrated.
16 For two straight wires, the final result is as abovea force inversely proportional to R, not to R 2.} There thus existed two kinds of forces associated with electricityelectric and magnetic. In 1864 James Clerk Maxwell demonstrated theoretically a subtle connection between the two types of force, unexpectedly involving the velocity of light. From this connection sprang the idea that light was an electric phenomenon, the discovery of radio waves, the theory of relativity and a great deal of presentday physics.
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