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I am working on a experiment on finding the Earth's magnetic field strength. Using a Galvanometer and a constant power source of 5.1V and slowly adjusting the resistance, I was able to find the changes in angle in relation to the changes of current.

https://en.wikipedia.org/wiki/Galvanometer#Theory

$$\frac{\mu_0nI}{2r}=B_H\tan\theta$$

Then I used the equation to find the $B_H$, the horizontal component BH of the Earth's magnetic field.

However, my calculation shows my $B_H$ is 5.62E-04, while the actual $B_H$ is 5.03E-05. There is a difference of factor of 10!!!

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  • $\begingroup$ Hard to say what it might be without actually being there and going over the setup and calculations in detail. Since, as you noted, the calculation is off by nearly a factor of 10 so I would start off by checking to make sure that there wasn't an accident or error in which a factor of 10 crept in. For example, N=50 turns but the spreadsheet says that the total coil is 500 turns. Sure that a partial and not total coil was used? Also, the magnetic inclination is a steep 70˚ in Denmark, so the field is far from horizontal there. Sure that the experiment was properly set up for that? $\endgroup$
    – user93237
    Sep 3, 2018 at 23:31
  • $\begingroup$ There are 2 settings on the device, 1 setting to turn on all 500 turns and 1 setting to turn on only 50 turns. I use the setting for 50 turns. If 500 turns is used in the calculation, the difference will be a factor of 100 instead of 10. μ (N/2r) I = Bh tan Bh = μ (N/2r) (1/slope) How did you get the 70˚ for the magnetic inclination? What would be a reasonable value? $\endgroup$
    – Tom
    Sep 4, 2018 at 6:14
  • $\begingroup$ I found out about the 70˚. Ignore that question:-) $\endgroup$
    – Tom
    Sep 4, 2018 at 6:23
  • $\begingroup$ There is possibly something inconsistent with your data. In your tabulated data to what does the resistance refer? Was it the resistance of a resistor in series with the tangent galvanometer? Was the supply voltage constant at $5.1\,\rm V$? If the answers are all yes then the extra resistance in the circuit (the resistance of the tangent galvanometer) varied from over $100 \Omega$ for low currents to below $30 \Omega$ for the higher currents. $\endgroup$
    – Farcher
    Sep 4, 2018 at 7:51
  • $\begingroup$ I will look into the volts/resistance/current issue by doing another experiment, if possible. I will take some photos next time so it is easier to double check the data. $\endgroup$
    – Tom
    Sep 4, 2018 at 10:40

2 Answers 2

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Maybe magnetic field at the point where you measured it included some local magnetic field, say, from some metal structure.

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It seems to be the case that such a galvanometer is most accurate at around $\theta=\pi/4=45^\circ$. The optimum current for this defection (assuming the correct $B_h$ is present is given by

$I=\frac{2rB_h}{\mu_0n}\approx1.2amps$

I wonder if it is correct to operate your particular device, as you seem to have done, with a deflection under $\theta=\pi/20=10^\circ$ and at currents an order of magnitude less than 1.2amps? (Note that incorporating more turns would also reduce the optimum current).

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  • $\begingroup$ Thanks for the feedback. Let me know what I can do with that info:-) $\endgroup$
    – Tom
    Sep 4, 2018 at 10:37
  • $\begingroup$ To use this you'ld probably have to run the experiment again with more turns and/or more current. Also run it a different location (to dis-count local ferromagnets) was another suggestion. $\endgroup$
    – JMLCarter
    Sep 4, 2018 at 16:39

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