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| 184_notes:examples:week9_detecting_b [2021/07/01 15:16] – schram45 | 184_notes:examples:week9_detecting_b [2021/07/05 21:58] (current) – [Solution] schram45 | ||
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| \vec{B}_3 &= \frac{\mu_0}{4 \pi}\frac{q\vec{v}\times \vec{r}_3}{{r_3}^3} = 4.2 \text{ nT } \hat{z} | \vec{B}_3 &= \frac{\mu_0}{4 \pi}\frac{q\vec{v}\times \vec{r}_3}{{r_3}^3} = 4.2 \text{ nT } \hat{z} | ||
| \end{align*} | \end{align*} | ||
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| + | Observation location 3 is the furthest away from our moving point charge and we would expect it to have a smaller magnetic field than location 2, this is reflected in our solution. We also expected the magnetic field at location 1 to be 0 since the velocity and separation vector are parallel for this point (this is always a good thing to look for when approaching a problem with cross products). | ||