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| 184_notes:examples:week9_earth_field [2021/07/01 15:30] – schram45 | 184_notes:examples:week9_earth_field [2021/07/05 21:55] (current) – [Solution] schram45 | ||
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| $$q=\frac{4 \pi h^2 B_{\text{earth}} \tan \theta}{\mu_0 v} = 131 \text{ C}$$ | $$q=\frac{4 \pi h^2 B_{\text{earth}} \tan \theta}{\mu_0 v} = 131 \text{ C}$$ | ||
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| + | This is a lot of charge. Since magnetic fields are very small in magnitude, and this can be seen in the magnetic constant, it makes sense that we would need such a large amount of charge to have a noticeable effect on our compass. Uf we were to compare the magnetic field of a moving charge to the electric field from that charge, the electric field would be much larger. | ||