Mutual Inductance Laboratory

Transcription

1 UNIERSITY OF ALABAMA Departent of Physics and Astronoy PH 106 Fall 2008 For this lab, you will need: 1 function generator 1 handheld ultieter 1 pair of coaxial solenoids (pictured below) 4 banana cables 1 ale BNC to banana plug adaptor 1 ruler or calipers Mutual Inductance Laboratory Introduction Consider two coaxial solenoids, one inside the other, a situation we discussed recently in lecture: Figure 1: Scheatic of your coaxial solenoids In the arrangeent shown at left, the outer coil is powered with an ac (tie-varying) voltage. The resulting tie-varying current creates a tie varying agnetic field, which penetrates the inner coil uniforly. Thus, there is a tie-varying agnetic flux in the inner coil, which induces a voltage on the inner coil. This voltage will also be tie-varying, with the sae frequency the voltage driving the outer coil. We derived an expression relating the rate at which the current in the outer coil changes with tie to the agnitude of the voltage induced across the inner coil: outer = M di inner dt (1) The constant of proportionality M is the utual inductance of the two coils. If the current in the

2 inner coil varies sinusoidally with frequency f, this expression becoes: outer = 2πfMI inner (2) Rearkably, the situation is exactly the sae if we instead power the inner coil and easure the voltage on the outer coil, as shown at right in the figure, even though the field produced by the shorter inner solenoid is terribly nonunifor. The relation between voltage in one coil and current in the other reains exactly the sae, no atter which is powered and which is easured. Therefore: we choose the easier syste to actually easure, naely, the right-hand one. Question: Why do we say the situation on the right in Figure 1 is easier to easure? For two solenoids, the utual inductance M ay be calculated exactly. Let the inner solenoid have length l and diaeter 2a, the outer solenoid diaeter 2b, as shown below 2a 2b l M = µ 0 πa 2 n inner n outer l Figure 2: Mutual inductance of two solenoids In general, utual inductance depends only on the geoetry and nuber of turns in each coil, as in this exaple. If the outer coil copletely encloses the inner, we can ake one ore siplification, as we did when discussing transforers: we know that the flux produced by the inner coil is copletely captured by the outer coil, so the product of the nuber of turns and current is the sae for both coils n inner I inner = n outer I outer (3) Using this relationship and the equations above, we can write the voltage on the outer coil as a function of the frequency driving the inner coil: outer = [ 2µ o π 2 a 2 li outer ] n 2 outer f (4)

3 The quantity in brackets depends only on constants, the geoetry of the inner coil, and the current in the outer coil, which you can deterine fro the easured voltage and resistance. i Thus, a plot of versus f should allow us to easure the nuber of turns in the outer coil. Preliinaries Measure the diaeter of both coils, and the length of the inner coil. Record these nubers below. Your coils should look soething like this: Figure 3: Your coaxial solenoids should look soething like this. Count the nuber of turns on your inner coil, and record this nuber below. Measure the resistance of the outer coil using the hand-held volteter. Record this nuber below. It should be Ω in ost cases. Procedure Turn on your function generator, and connect its output ii to the handheld volteter. Set the function generator s frequency to f = 1 khz, its wavefor to a sine wave, and axiize the aplitude. With the volteter in ac volts ode (sybolized by ), easure the output of the function generator and record it. Now connect the function generator output to the inner coil instead of the volteter. This will power the inner solenoid. Connect the outer coil s terinals to the volteter, which should still be in ac volts ode (sybolized by ) Place the inner solenoid copletely inside the outer one, and note the voltage easured on the outer coil. Question: What happens when you ove the inner coil slowly out of the outer coil? Why? Replace the inner coil such that is copletely enclosed by the outer coil. i Since n inner I inner = n outeri outer, we can deterine the current at a single frequency, and presue it constant over the course of the easureent. The voltage on the outer coil will change, but its current will be roughly the sae over the frequency range of interest. ii Use the 50 Ω output, not the TTL" output.

4 In steps of f = 1 khz, record the voltage on the outer coil versus the frequency of the voltage applied to the inner coil fro 1 10 khz using the table below. Do not change the aplitude of the function generator output applied to the inner coil. Plot the voltage on the outer coil versus frequency of the inner coil s signal in excel, as shown in the exaple below. Using a linear trendline, deterine the slope of the curve and record it below.? slope = 1.73 x 10-4 /Hz outer f (Hz) Figure 4: Exaple plot of the outer coil s voltage versus the frequency of the voltage applied to the inner coil. Using the relation below, deterine the nuber of turns in the outer coil, and record this below. Deterine I outer fro the easured resistance, and the easured voltage at 1 khz. Since we easure only the agnitude of the voltage, the inus sign is irrelevant here. outer = [ 2µ o π 2 a 2 li outer ] n 2 outer f = n 2 outer = slope [2µ o π 2 a 2 li outer ] (5) Is your value reasonable? Geoetrically estiate (roughly) the nuber of turns per layer and the nuber of layers in your outer coil to verify. When you are finished Power off the function generator and ultieter Straighten up your coponents and wires, return the to the cart Turn in a hard copy of your report and plots

5 Data f (khz) 1.0 outer coil voltage () function generator output o frequency outer coil resistance R o outer coil current I o = o /R o turns on inner coil n inner inner coil diaeter 2a inner coil length l outer coil diaeter 2b Hz Ω A