Helmholtz Coils and Magnetic Fields PDF

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2EM7 Magnetic Fields of Helmholtz Coils By Akmal Hafizi Annuar (148208) December 2020 Second Year Laboratory Report ZCT 293/2 1 2EM7 Magnetic Fields of Helmholtz Coils ABSTRACT The objectives of the experiment are to determine the magnetic field along the horizontal x-axis that passes through the ce...

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2EM7 Magnetic Fields of Helmholtz Coils

By

Akmal Hafizi Annuar (148208)

December 2020

Second Year Laboratory Report ZCT 293/2

1

2EM7 Magnetic Fields of Helmholtz Coils ABSTRACT The objectives of the experiment are to determine the magnetic field along the horizontal x-axis that passes through the centre of a single solenoid coil, and to determine the magnetic field along the horizontal xaxis that passes through the centre of the Helmholtz coil. Helmholtz coil is a device that produces a region of a nearly uniform magnetic field. It consists of two solenoids that are parallel to each other on the same axis. Both solenoids are separated by a distance, d. Each coil carries an equal electric current in the same direction. The entire experiment is conducted via a simulator software provided. For Experiment I, the graph of B vs x is obtained alongside with the logarithmic graph of B vs the square of x. The comparison of the experimental and the theoretical logarithmic graphs allows the determination of the turns of wire, N of the hypothetical single coil. That is, N = 1717.5. It is managed to obtain the best value for B0 through the standard deviation as the uncertainty in a single measurement with 70% confidence. That is, B0 = (4.1267 x 10-3) ± (9.2236 x 10-5) T. The experimental µ0 is deduced and it is given by µ0 = (2.5292 x 10-7) T m A-1. The determination of the experimental µ0 yields a percentage error of 79.9%. For Experiment II, the graph of B vs x is obtained for all d = R, d = 1.5R and d = 0.5R. Two major things found out in this part are, firstly, the mathematical erratum in either the simulator or in the laboratory manual is very substantial, and secondly, the erratum has caused such an ambiguity that a thorough quantitative analysis has become cumbersome given the time constraint as the deviation between the experimental and the theoretical values are of logarithmic. Next, the graph of B0 vs d is also obtained for both the experimental and the theoretical values. Nothing much could be done on the quantitative aspect of it. However, qualitatively, it is observed that as d increases, B decreases. This may explain the lesser incident flux density as the coils move further apart. Lastly, the slope of the experimental data has a greater rate of change as opposed to that of the theoretical values.

2

THEORY 1. Introduction The Helmholtz coil is a device that produces a region of a nearly uniform magnetic field. It consists of two solenoids that are parallel to each other on the same axis. Both solenoids are separated by a distance, d, which is equal to the radius, R = 0.105 m, of the coil. Each coil carries an equal electric current in the same direction. The magnetic field along the x-axis that passes through the center of the solenoid(s) is to be measured [3]. 2. Theory Single Coil For a coil of wire having radius R and N turns of wire, the magnetic field along the perpendicular axis through the center of the coil (x is the distance from the center of the coil) is given by [1] 𝐵=

𝜇0 𝑁𝐼𝑅2

2(𝑥 2

+

3 𝑅 2 )2

(1)

The magnetic field at the center of the coil is obtained by setting x = 0, to give [1] 𝐵0 =

𝜇0 𝑁𝐼 2𝑅

Figure 1 shows the single coil in theory [1].

Figure 1. Single coil.

3

(2)

Two Coils Figure 2 shows two coils with an arbitrary separation [1].

Figure 2. Two coils with an arbitrary separation. For two coils, the total magnetic field is the sum of the magnetic fields from each of the coil [1]. It follows that 𝐵� = 𝐵�1 + 𝐵�2 =

𝜇0 𝑁𝐼𝑅2 2

𝑑 �� − 𝑥� + 𝑅2 � 2

� 3𝑥 2

+

𝜇0 𝑁𝐼𝑅2 2

𝑑 �� + 𝑥� + 𝑅2 � 2

� 3𝑥 2

(3)

For Helmholtz coils, the coil separation (d) equals the radius (R) of the coils. This coil separation gives a uniform magnetic field between the coils. Substituting x = 0 gives the magnetic field at a point on the xaxis centered between the two coils [1], 𝐵� =

8𝜇0 𝑁𝐼

√125𝑅

𝑥�

Figure 3 shows the Helmholtz coils [1].

Figure 3. Helmholtz coils.

4

(4)

EXPERIMENTAL METHODOLOGY The experiment starts with having the DC supply switched on. The slider of the DC supply is adjusted to achieve the desired coil current. Then, the magnetometer is switched on [2]. For the double coils experiment, the horizontal position of the coils is adjusted by moving the sliders of both coils. The horizontal position of the magnetic field sensor is reset. The REC button is clicked to begin the recording of the magnetic field, B against the horizontal position, x. The magnetic field sensor is slid to make it goes through the center of the coil(s) until the other end of the horizontal axis. Finally, the STOP button is clicked to end the reading [2]. Next, the “Data” tab is chose and the “Copy All” button is clicked to copy the recorded data to the clipboard. The copied data is pasted into an Excel sheet to create a scatter plot of B against x [3]. Experiment I Experiment I is to determine the magnetic field along the horizontal xaxis that passes through the centre of a single solenoid coil. The “Single Coil” tab is clicked to display the virtual apparatus setup of Experiment I [3]. The graph of B against x is plotted. The graph of ln B against ln (x2 + R2) is plotted. The gradient of the log-log graph is deduced. The theoretical curve of ln B against ln (x2 + R2) based on Equation (1) is plotted on the same log-log graph. The expected theoretical value of the gradient is obtained. The measured value of the gradient is compared with the theoretical value. The value of N of a single Helmholtz coil is deduced. The given value of N is then compared and the accuracy of the measurement of N is discussed. From the graph obtained, the experimental value of B0 is deduced. Using the values of the measured I, N and R, the value of µ0 is deduced. The theoretical value of µ0 is then compared. The results of the measurements are discussed [1].

5

Figure 4. The single coil experiment. Experiment II Experiment II is to determine the magnetic field along the horizontal xaxis that passes through the centre of the Helmholtz coil. The "Double Coils“ is clicked to display the virtual apparatus setup of Experiment II [3]. For each coil-coil separation distance, d, a graph of B against x, and a graph of the theoretical curve of Equation (3), is plotted on the same graph. The graph obtained is discussed, particularly at the region of which the magnetic field remained constant. From the graph obtained, a graph of experimental B0 against theoretical B0 is plotted on the same graph. B0 is the value of the magnetic field at the midpoint between the Helmholtz coils. The results of the measurements are discussed, particularly the factors that caused the discrepancies [1].

6

Figure 5. The double coils experiment.

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EXPERIMENTAL DATA Since the data involved are of hundreds and for convenience, all of the experimental data of Experiment I can be referred to Appendix A. Likewise, all of the experimental data of Experiment II can be referred to Appendix B.

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DATA ANALYSIS & DISCUSSION Experiment I: Single Coil Experiment I is set up such that the radius of the coil, R = 0.105 m, the electric current, I = 1.995 A, the recording rate is 10 Hz and the magnetometer is of 103 Tesla. Firstly, Figure 6 shows the graph of magnetic field, B against displacement, x, where the measurement of the displacement is of the manner from left-hand side to right-hand side.

Graph of B (kT) vs x (cm)

-50

-40

-30

-20

-10

4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 -0.5 0

10

20

30

40

50

Figure 6. Graph of B vs x from LHS to RHS. It follows that, Figure 7 shows the graph of magnetic field, B against displacement, x, where the measurement of the displacement is of the manner from right-hand side to left-hand side.

9

Graph of B (kT) vs x (cm)

-50

-40

-30

-20

-10

4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 -0.5 0

10

20

30

40

50

Figure 7. Graph of B vs x from RHS to LHS. Now, since there are technically two data collections for the same graph of B against x, then for the best estimate for the quantities, we calculate the average value of the magnetic field, B and the displacement, x. It is worth noting that the calculation for the mean value is not quite straightforward. Firstly, both measurements for x are basically opposite in polarity to each other. For instance, in reference to Appendix A3, one end of the first measurement is -35, and hence the same end of the second measurement is 35. So, we calculate the absolute value of both measurements of x. Then, from the absolute values, then only we calculate the average value of the displacement, x. Lastly, from the innate sequence of the average value of x, we turn those values that come prior to null (zero) into negative integers, to retain the nature of the experiment. Likewise, the same mechanism is applied to the determination of the average of value of the magnetic field, B. All in all, Appendix A3 best tabulates the above description in the determination of the mean value for B and x. With that, Figure 8 shows the graph of averaged-B against averaged-x. For the convenience of the subsequent analysis, we can denote the average value of B and x as just mere B and x.

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Graph of averaged-B (kT) vs averaged-x (cm) 4 3.5 3 2.5 2 1.5 1 0.5 0 -50

-40

-30

-20

-10

0

10

20

30

40

-0.5

Figure 8. Graph of the average value of B against of x. Essentially, Figure 8 shows the desired graph of B vs x. Referring to the graph, the curve is all in all reasonably close to the desired one. What matters, as what is achieved from the graph, is the seemingly bellshaped normal distribution curve. This is for the reason that it is expected to measure a higher reading of magnetic field near the null (zero) displacement than that at both ends of the curve. Also, it is for the reason that the nature of the experiment is symmetrical in the context of the reading of the magnetic field as a function of the displacement. Moreover, the graph shows that the standard error is appreciably quite large. But it is reasonable because the graph of Figure 8 is actually the mean value of two measurements of the same procedure that made in the simulator. The plus side of it, is we obtain a better estimate for both B and x readings. Lastly, we can clearly see the whole curve is shifted a bit to the positive x-direction, as opposed to the desired maximum point to be exactly at x = 0. This may be due to the systematic error rooting from the codes in the simulator software being amplified due to the mechanical calculation of the mean value being done to the initial two measurements made. Also, it may be due to the random error made when conducting the slider in the simulator.

11

Moving on to the next procedure assigned, Figure 9 shows the graph of ln B vs ln (x2+R2).

Graph of ln B vs ln [x^(2)+R^(2)] 2

1

0 -5

-4.5

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0 -1

-2 y = -1.5017x - 5.3359 -3

-4

Figure 9. Graph of ln B vs ln (x2+R2). The gradient of the log graph is computationally obtained and it is such that ln 𝐵 = −1.5017[ln(𝑥 2 + 𝑅2 )] − 5.3359

(5)

It is obtained that the curve has a negative gradient and its y-intercept is about -5. Now, it is worth to note that the graph obtained computationally is such that the readings obtained for x is first changed its units from cm into m. Then, the square of x is calculated. Subsequently, the sum of xsquared and R-squared is obtained. Lastly, the log of the sum is calculated. Likewise, the log of B is calculated. Appendix A4 best tabulates the computation involved in the excel sheet as described above in the determination of the log of both x and B. Next, we plot on the same graph the theoretical curve of ln B vs ln (x2+R2) based on the Equation (1). Figure 10 shows the theoretical graph of ln B

12

vs ln (x2+R2) from Equation (1) together with the experimental graph (Figure 9) obtained from the measurements.

Graph of ln B vs ln (x^2+R^2) 5

0 -5

-4

-3

-2

-1

0 -5 Theoretical -10

Experimental Linear (Theoretical)

-15

-20

y = -1.5x - 23.159

-25

Figure 10. Comparison of theoretical and experimental graph of ln B vs ln (x2+R2). The gradient of the theoretical log graph is computationally obtained and it is such that ln 𝐵 = −1.5[ln(𝑥 2 + 𝑅2 )] − 23.159

(6)

It is obtained that the curve also has a negative gradient and its yintercept is about -23. As mentioned, the mechanism in determining the theoretical graph is such that the Equation (1) is used to calculate the theoretical ln B and ln (x2+R2). Most importantly, the assumptions made in the determination of the theoretical values via Equation (1) are N = 1, R = 0.105 m, I = 1.995 A, and 𝜇0 = 4𝜋 × 10−7 . Appendix A5 best tabulates all the data obtained experimentally and the data theoretically calculated in the determination of the comparison of both log curves. Now, to compare, since we let N = 1 in our calculation using Equation (1), then we expect and it is observed that the theoretical curve to has its ln B being smaller in values as opposed to the experimental measurement.

13

As a side note, the assumption of N = 1 is part of the motive to deducing N in the next procedure. Accordingly, it is obtained that the y-intercept of the experimental data and the theoretical calculation is roughly -5 and 23, respectively. Again, it is expected of the theoretical value to be of smaller than that of the experimental data. It is worth to note that both the curves have approximately the same gradient that is about -1.5. Accordingly, the only difference between the two curves is the y-position in the graph of Figure 10 and the assumptions made rooting to that is for the deduction of the experimental N. Moving on to the next assigned procedure, deduction of N turns out to be more straightforward than it initially seemed. We start by comparing both experimental and theoretical magnetic field readings, B. However, in order to deal with B, we must first determine the value of x, both of experiment and theory. Having said that, for convenience, let assume x = R = 0.105 m. Next, we need to obtain the corresponding experimental B when x = R = 0.105 m. So, when it comes to experimental data, we can now rely on Equation (5) for the determination of the corresponding experimental B. From Equation (5), it follows that ln 𝐵 = −1.5017[ln(𝑥 2 + 𝑅2 )] − 5.3359

It is important to note that the x and R is of cm unit and the B is of kT unit, so the former should become 10.5 cm prior to substituting into Equation (5). It follows that ln 𝐵 = −1.5017[ln([10.5]2 + [10.5]2 )] − 5.3359 𝑒 ln 𝐵 = 𝑒 −16.8967

Thus, the experimental B, when x = R = 0.105 m, is given by 𝐵 = 4.5905 × 10−5 𝑇

Now, we use Equation (1) to obtain the number of turns of wire, N for the hypothetical single coil in the simulator. From Equation (1), it follows that

14

𝐵=

𝜇0 𝑁𝐼𝑅2

3

2(𝑥 2 + 𝑅2 )2

Substituting the experimental B at x = 0.105 m, µ0, I = 1.995 A, R = x = 0.105 m,

4.5905 × 10−5

1 × 10−7 � (1.995)(0.105)2 = 𝑁 4𝜋 3 2(0.1052 + 0.1052 )2 �

Thus, the turns of wire, N of the hypothetical single coil in the simulator is given by 𝑁 = 1717.5 ≈ 1718

Therefore, it is approximated that the hypothetical single coil in the simulator has 1718 turns of wire. Now that we have obtained the experimental N, note that during the process we obtained the corresponding experimental B whose value is of 10-5 T. Whereas, the measurement taken for B in the simulator is of 103 T. This is one of the oddness detected of which the source of error are failed to be pin-pointed. Nevertheless, we may assume the particular error is of an overlooked systematic error somewhere in the entire process. Moving on to the next procedure assigned, we deduce the experimental value of B0, the magnitude of B when x = 0. From Equation (5), it follows that

Let x = 0,

ln 𝐵 = −1.5017[ln(𝑥 2 + 𝑅2 )] − 5.3359 ln 𝐵 = −1.5017[ln(0 + [10.5]2 )] − 5.3359 𝑒 ln 𝐵 = 𝑒 −12.3980 15

Thus, B0 is given by, 𝐵0 = 𝐵 = 4.1267 × 10−6 𝑘𝑇 = 4.1267 × 10−3 𝑇

Therefore, it is obtained that the experimental value of B0, the magnitude of B when x = 0, is (4.1267 x 10-3) T. Now, we know that the reading of B in the simulator is of 103 T. And here we obtain the calculated B from Equation (5) is of 10-3 T. In virtue of calculating the uncertainty, let assume there is an overlooked error made in the simulator software by which it instead meant 10-3 when it apparently stated 103. Having said that, we may further obtain the best value for B0, the magnitude of B when x = 0. To obtain the best value for B0, we have to apply the standard deviation as the uncertainty in a single measurement [4]. Where, it stated that the best value for the second reading can be of the form such that its mean value is of the second reading but its standard deviation is of the first reading, all of which with 70% confidence [4]. Equation (7) best sums up the definition above [4], 𝑥2 = 𝑥̅2 ± 𝑆(𝑥1 )

(7)

With that, let the 𝑥̅2 be the obtained B0 = (4.1267 x 10-3) T. Then, what is left to calculate is the S(x1). To start, we have to first obtain the highest value of B from the measurements made in Appendix A1 and Appendix A2, under the assumption that these maximum values of B are the closest to the experimental displacement, x at null, that is x ≈ 0. Table 1 shows the calculation involving the readings of maximum B for the standard deviation [4]. Table 1. The calculation of the standard deviation. Measurement Appendix A1 Appendix A2

Value, Bi 3.84441 3.857992

Deviation, di = Bi - 𝐵� -6.791 x 10-3 6.791 x 10-3

16

di2 4.6118 x 10-5 4.6118 x 10-5

� 𝑑𝑖2

𝐵� = 3.851201

= 9.2236 × 10−5

It follows that, the equation of standard deviation is given by [4], 𝑁

It follows that,

1 𝑆 (𝐵 ) = � �[(𝐵𝑖 − 𝐵�)2 ] 𝑁−1

(8)

𝑖=1

1 (9.2236 × 10−5 ) = 9.2236 × 10−5 𝑇 𝑆(𝐵0 ) = 𝑆(𝐵) = � 2−1

Therefore, the best value for the experimental B0, with 70% confidence, is given by, best value of B0 = (4.1267 x 10-3) ± (9.2236 x 10-5) T From the best value of B0 obtained, it is obviously a very minute discrepancy and this may be due to the nature of the experiment whereby it is conducted using a pre-programmed simulator software. Thus, the alleged computation gives rise to the tiny corresponding discrepancies. Moving on to the next assigned procedure, using the given values, I = 1.995 A, and R = 0.105 m, and the obtained values, N = 1717.5, and B0 = (4.1267 x 10-3) ± (9.2236 x 10-5) T, the experimental µ0 is to be deduced. Expressing for µ0 from Equation (2), yields 2𝐵0 𝑅 2(4.1267 × 10−3 )(0.105) 𝜇0 = = = 2.5292 × 10−7 𝑇 ∙ 𝑚 𝐴−1 (1717.5)(1.995) 𝑁𝐼

Therefore, it is obtained that the...