# Experiment A5. Hysteresis in Magnetic Materials

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1 HYSTERESIS IN MAGNETIC MATERIALS A5 1 Experiment A5. Hysteresis in Magnetic Materials Objectives This experiment illustrates energy losses in a transformer by using hysteresis curves. The difference betwen the B and H fields encountered in the lecture courses is illustrated. You will: use a hysteresis curve to measure the power loss of an iron core transformer for comparison, measure the loss for a ferrite core transformer estimate the curie point for ferrite. Prework Questions. 1. Briefly explain what the physical meanings are of the three magnetic quantities (B, H, and M) mentioned in the Background. 2. How is the hysteresis curve related to the efficiency of a given transformer? 3. Draw what you expect the hysteresis curve in Fig. A5-3 would look like (paying attention to the position of the intercepts H c and B R ) for (a) a larger amplitude and (b) a smaller amplitude of the AC current, with the temperature and frequency unchanged. Topic 1. Magnetisation and Hysteresis Curves Background In the context of magnetic materials, hysteresis refers to the following phenomenon: When a piece of unmagnetised iron is placed in a solenoid which is carrying a current, the iron will become magnetised. If the current in the solenoid is then reduced to zero, the iron will remain partly magnetised. In order to demagnetise the iron, it is necessary to reverse the direction of current in the solenoid, but if the current in the reverse direction is increased further, the iron will magnetise in the reverse direction. A model made of small compass needles is available on a side bench to demonstrate these effects.

2 A5 2 GROUP A EXPERIMENTS i B ~ H ~ Fig. A5-1 B and H inside an iron core To distinguish between (a) the total magnetic field in the iron (i.e., the field due to the solenoid plus the field due to the magnetisation of the iron), and (b) the part of the magnetic field created by the solenoid, it is useful to define two magnetic field quantities: one is the magnetic field, B, the other is an auxiliary field H. These quantities are often referred to simply as the B field and the H field. If we wind a coil of N turns on an iron ring and pass a current i through the coil, then the value of H in the iron can be found from Ampere s law H dl = Ni, giving H = Ni/l, (1) where l is the length of the iron ring (i.e., the length of the dotted line in Fig. A5-1). The value of H is proportional to i and N but does not depend on the state of magnetisation of the iron. The iron itself produces a magnetic field which depends on the degree of alignment of magnetic dipoles in the iron set up by orbital and spinning electrons. The degree of alignment can be specified by a quantity known as the magnetisation, M. In an unmagnetised piece of iron, M = 0, but the dipoles remain aligned over small regions in the iron known as magnetic domains. In an unmagnetised piece of iron, the domains are aligned in random directions. The effect of applying an H field is to align some of the domains to produce a non-zero value of M. The total magnetic field in the iron is B = µ 0 (H + M), (2) where µ 0 is the permeability of free space. Experimentally, we can measure B by winding a second coil on the iron core and measuring the emf induced in the coil when B changes with time. As the current i in Fig. A5-1 increases from zero, the magnetisation M will increase from zero up to a certain value at which all the domains in the iron are perfectly aligned. Any further increase in i will have no effect on the value of M, and the iron is said to be saturated. A curve of the total field B vs. the applied field H as H increases from zero is called a magnetisation curve. A typical magnetisation curve for iron is given in Fig. A5-2. Suppose we reach some arbitrary point (H 0, B 0 ) on the magnetisation curve shown (dotted) in Fig. A5-3. If we then decrease H to zero (by decreasing the current in the external coil), the iron will remain partly magnetised and there will be a residual field B R (see Fig. A5-3). By reversing the current, we can decrease B to zero at a value of H known as the coercive

3 HYSTERESIS IN MAGNETIC MATERIALS A5 3 Fig. A5-2 Magnetisation curve Fig. A5-3 Hysteresis curve force, H c. As H is made more negative, the iron magnetises in the reverse direction, and it will arrive at the point ( H 0, B 0 ) when the reverse current is equal in magnitude to the initial forward current. A curve of B vs. H for a complete cycle of increasing and decreasing current is known as a hysteresis curve. Various hysteresis curves are possible for a given specimen of iron, depending at which point on the magnetisation curve the hysteresis curve is started. Apparatus N = 100 N =30 HI (red) i Oscilloscope Y Channel ~ LO (blue) A R = 5Ω Transformer v s ~ Integrator vo Oscilloscope Earth Oscilloscope Earth Oscilloscope X Channel Fig. A5-4 Circuit used to display hysteresis curves The circuit shown in Fig. A5-4 can be used to display the hysteresis curves of an iron core transformer directly on an oscilloscope screen. The transformer consists of a primary coil of N 1 = 100 turns and a secondary coil of N 2 = 30 turns, both wound on an iron core. A high power oscillator is used to produce an alternating current i in the primary coil. The H field in the iron is related to the current by eqn. (1). The current is indicated on a peak-reading ammeter A, which indicates the amplitude of the current. The current amplitude can be varied by a gain control on the oscillator, and the alternating frequency can be varied from 1 Hz to 30 khz. The magnetic field B in the iron varies with time and induces a voltage v s = N 2 A (db/dt) (3)

4 A5 4 GROUP A EXPERIMENTS in the secondary coil, where A is the cross-sectional area of the iron. Since we need to measure B rather than db/dt, we will integrate the voltage v s electronically. The basic principles of integrating circuits are described in the circuits and electronics notes. The battery operated integrator in Fig. A5-4 produces an output voltage v o which is related to the input voltage v s by v o = 1 RC = N 2A RC v s dt db dt dt = N 2A B, (4) RC where R and C are the resistance and capacitance, respectively, of the components used in the integrating circuit. The integrator provided has R = 1 MΩ and a gain control switch to select either C = 0.01 µf or C = µf. For uncertainty calculations, assume tolerances of 2% on R and C. Both B and v o will vary with time, but at all times B = RC N 2 A v o. (5) The output signal from the integrator is connected to the Y input of an oscilloscope to produce a vertical deflection of the oscilloscope trace. The vertical deflection formed on the oscilloscope screen depends on the applied voltage v o and on the VOLTS/CM setting on the oscilloscope, and is proportional to B. The ammeter used to measure the primary current has a resistance of R A = 5.0 Ω, so the voltage across the meter will be R A i. This voltage signal can be applied to the X input of the oscilloscope to produce a horizontal deflection of the oscilloscope trace proportional to i, and therefore to H. When conducting the experiment, take care not to let the current rise above about 1.2 A because the fuse in the ammeter will burn out. The oscilloscope trace will therefore deflect vertically by an amount proportional to B and horizontally by an amount proportional to H. As the current in the primary coil alternates between positive and negative values, a hysteresis curve (B vs. H) on the oscilloscope screen will automatically be traced out. Procedure 1. Connect the circuit shown in Fig. A5-4. In this experiment we will use an oscilloscope as an X Y display to analyse the hysteresis curve. Set the coupling for each channel on the oscilloscope to DC. If your hysteresis curve appears backwards, reverse the leads connecting the transformer to the integrator. 2. Set the gain control switch on the integrator to C = 0.01 µf and obtain a hysteresis curve at an oscillator frequency f = 50 Hz. (If the curve on the oscilloscope display is highly distorted, then use the C = µf capacitor). Note that, as the amplitude

5 HYSTERESIS IN MAGNETIC MATERIALS A5 5 C1 of the primary current increases, the hysteresis curves grow in size. The extreme tips of the curves lie on the magnetisation curve (Fig. A5-3). Record the coordinates (H 0, B 0 ) of the tips of the hysteresis curves for primary currents between 0.05 A and 1.0 A. The coordinates can initially be recorded in volts (corresponding to the Channel 1 and Channel 2 V/cm settings). Plot the magnetisation curve (B 0 vs. H 0 ) on a linear graph, after converting your measured values to B (Tesla) and H (Amps metres 1 ) units using eqn. (1) and eqn. (5). The dimensions of the iron core are l = 125 mm and A = 19 mm 17.5 mm, with uncertainties of ±2% in both l and A. 3. The ratio B 0 /µ 0 H 0 = µ r for points along the magnetisation curve is called the relative permeability of the iron. For a vacuum, B = µ 0 H (since M = 0), so µ r is the factor by which B is increased due to the presence of iron. Use your results to plot a graph of µ r vs. H and compare your results with those given in the bench notes. 4. For current amplitudes i = 0.1 A and 1.0 A, make graphs of the hysteresis curves observed on the oscilloscope, labelling the axes carefully in B and H units. Curves such as these are used in studying the behaviour of transformers. The area of the hysteresis curve is important since it represents the work done in one hysteresis cycle per unit volume of iron. To prove this result, note that the current in the primary coil is i = Hl/N and the voltage across the primary coil is v = N 1 A (db/dt). Since the power used is p = vi, and the work done in a small time dt is dw = p dt, the total work done in one complete cycle is W = p dt = NA db Hl dt N dt = V H db. (6) Here, V = Al is the volume of the iron core and H db is the area enclosed by the hysteresis curve (in units of H B). Estimate the area enclosed by the 0.1 A hysteresis curve and hence estimate the power dissipated in the transformer at 50 Hz. Power = (Area of curve) (Volume of iron) (Frequency). C2 Topic 2. Eddy Currents The changing magnetic flux in the iron core of a transformer will induce an emf, not only in the primary and secondary turns, but also in the iron core. The iron core is a good conductor, so the currents induced in a solid iron core will be large. These currents are known as eddy currents and are undesirable, since they heat the core and result in power losses (in addition to the hysteresis losses). Furthermore, the eddy currents flow in a direction which, by Lenz s law, acts to weaken the flux created by the primary coil. Consequently, the current in the

6 A5 6 GROUP A EXPERIMENTS primary coil required to produce a given B field is increased, so the hysteresis curves are fatter along the H axis. The magnitude of the eddy currents is proportional to the operating frequency, since the induced voltage in the iron is proportional to the rate of change of flux. At high frequencies, it becomes impractical to use iron cores in transformers and it is necessary to use either air cores or ferrite cores. Ferrites (as used in transistor radio antennas) are strongly magnetic and behave almost as insulators, so the induced eddy currents are very small. Even at 50 Hz it is impractical to use solid iron cores in transformers or inductors. In practice, it is necessary to construct the core from thin strips of iron, known as laminations, which are coated with varnish to provide good electrical insulation. The eddy current paths are then broken (see Fig. A5-5), but smaller currents do flow within each lamination and limit the operating frequency to about 1 khz. The transformer used in this experiment is made from a stack of 39 such laminations (they are easily visible). Eddy Current CORE CORE Primary Current Primary Current Fig. A5-5 Eddy current paths are broken by laminating an iron core Procedure 1. To observe the effect of eddy currents, connect the circuit shown in Fig. A5-4, but use the N = 30 turn coil as the primary coil and the N = 100 turn coil as the secondary. This will allow you to saturate the core at frequencies up to about 500 Hz. 2. Because the impedance of the circuit changes with frequency, the current will vary as you change the frequency. You will need to adjust the power supply voltage as the frequency changes. Remember to keep the current below 1.2 A. Increase the oscillator frequency from 10 Hz to 500 Hz and note the fattening of the hysteresis curve at the higher frequencies. The width of the hysteresis curve can be measured in terms of the coercive force H c (see Fig. A5-3). Plot a graph of H c vs. frequency, keeping the amplitude of the primary current fixed at 1.0 A. Explain why the curves get fatter. 3. A ferrite transformer with N 1 primary turns and N 2 secondary turns has been placed in an oven with connection terminals outside the oven. The oven will be used in the next topic to examine the effect of heating the ferrite. The ferrite core has dimensions

7 HYSTERESIS IN MAGNETIC MATERIALS A5 7 of cross-sectional area A and length l. The values of N 1, N 2, A and l are marked on the experimental apparatus. Using the same circuit as Fig. A5-4, examine the ferrite hysteresis curve for a primary current of amplitude 1.0 A and at frequencies over the range 50 Hz < f < 10 khz. Note that the curve fattens slightly at high frequencies, but not nearly as much as the laminated iron core. Plot a graph of H c vs. frequency. Record the coordinates (in T and A m 1 ) of the tip of the curve at f = 1 khz. This coordinate can be plotted as the room temperature point for the next topic. C3 Topic 3. Curie Point for Ferrite When magnetic materials are heated, thermal energy destroys the ability of magnetic domains to align along an external magnetic field. As the temperature rises, the magnetisation decreases until a temperature, called the Curie point, is reached at which M becomes very small. Procedure 1. Reconnect the setup by replacing the transformer with the ferrite in the oven. The temperature is recorded by means of an electronic thermometer; the sensor is a thermocouple which is in good thermal contact with the ferrite which is in the oven. Record the current temperature (which should be near room temperature,), then switch on the oven. 2. With f = 1 khz and i = 1.0 A, sketch roughly the hysteresis curve at various temperatures as the temperature rises. 3. When the temperature reads 100 C, switch off the oven. You will observe that even though the oven is switched off, the temperature will continue to rise to around 130 C, and then begin to fall. As the temperature falls, record the vertical amplitude of the tip of the hysteresis curve. Since the relative permeability changes rapidly near the Curie point, make sure you record enough points in this region. Plot a graph of relative permeability µ r (= B 0 /µ 0 H 0 ) as a function of temperature. (The value of µ r for this sample refers to a value of H = N 1 i/l. The dimensions of the ferrite transformer are given in the previous topic.) You should find that µ r decreases with increasing temperature towards the vacuum value µ r = 1.0. It never gets as low as 1.0, but we can extrapolate the steepest part of the µ r vs. T curve towards B = 0 and define the intersection with the T axis as the Curie point. Use your results to estimate the Curie point for ferrite.

8 A5 8 GROUP A EXPERIMENTS Conclusion C4 Comment on the significance of your observations in this experiment. For example, how much benefit does a ferrite core promise compared to an iron core? What is the significance of the Curie point to transformer operation?

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