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--- 184_notes:examples:week9_earth_field [2017/10/20 14:14] – tallpaul
+++ 184_notes:examples:week9_earth_field [2021/07/05 21:55] (current) – [Solution] schram45
@@ Line 1: / Line 1: @@
+[[184_notes:perm_mag|Return to Permanent Magnet notes]]
 =====Using the Earth's Magnetic Field for Measurements=====
-You have spotted an unidentified flying object! Naturally, you wish to find its charge. You have a compass, a good sense of direction, and keen eyesight. You notice that it is flying due south on a course that will pass directly overhead, and it is $30 \text{ m} $above you, travelling at$ 200 \text{ m/s}$. You observe the dial on your properly aligned compass as the object passes overhead. As a function of time, this is what you see:
+You have spotted an unidentified flying object! Naturally, you wish to find its charge. You have a compass, a good sense of direction, and keen eyesight. You notice that it is flying due south on a course that will pass directly overhead, and it is $15 \text{ m} $above you, travelling at$ 200 \text{ m/s}$. You observe the dial on your properly aligned compass as the object passes overhead. As a function of time, this is what you see:
-{{ 184_notes:9_compass_graph.png?400 |Compass and Theta Graph}}
+[{{ 184_notes:9_compass_graph.png?500 |Compass and Theta Graph}}]
 ===Facts===
-  * We know $|\vec{B}_{\text{earth}}| = 32 \mu\text{T}$.
+  * We know $B_{\text{earth}} = 32 \mu\text{T}$.
-  * If we align our coordinate axes, according to the represention below,  $\vec{B}_{\text{earth}} = 32 \mu\text{T } \hat{y}$ .
+  * If we align our coordinate axes according to the representation below,  $\vec{B}_{\text{earth}} = 32 \mu\text{T } \hat{y}$ .
-  * $h = 30 \text{ m}$.
+  * $h = 15 \text{ m}$.
   *  $\vec{v} = -200 \text{ m/s } \hat{y}$ .
   * You have the graph of  $\theta$  versus  $t$ .
 ===Lacking===
-  *  $\vec{B}_{text{UFO}}$
+  * $\vec{B}_{\text{UFO}}$
+  * $q$
 ===Approximations & Assumptions===
-  * The UFO can be approximated as a moving point charge.
+  *  $q$ ,  $\vec{v}$ ,  $h$ , and  $\vec{B}_{\text{earth}}$  are all constants: All these variable could easily be changing in this problem. However, to simplify down the model to better analyze the situation we will assume all these things are constant. This could be an accurate assumption if the UFO were cruising steadily over us at the given height.
-  *  $q$ ,  $\vec{v}$ ,  $h$ , and  $\vec{B}_{\text{earth}}$  are all constants.
   * Your sense of direction and eyesight can be trusted.
@@ Line 22: / Line 24: @@
   * We represent the Biot-Savart Law for the magnetic field from a moving point charge as
  $\vec{B}= \frac{\mu_0}{4 \pi}\frac{q \vec{v}\times \vec{r}}{r^3}$
+<WRAP TIP>
+===Approximation===
+We will approximate the UFO as a point charge. This may not be the most accurate assumption since the UFO is most likely a large object, but it is an approximation that must be made in order to apply our knowledge to the problem. Since the problem doesn't state anything otherwise and it is a necessary assumption to use the equation above, we will approximate it as a point charge.
+</WRAP>
   * We represent the situation with the following pictures. Coordinate axes and cardinal directions are specified.
-{{picture}}
-{{picture}}
+[{{ 184_notes:9_ufo_representation.png?400 |A Glimpse of the Unknown}}]
+[{{ 184_notes:9_compass_coordinates.png?300 |Compass}}]
 ====Solution====
-Below, we show a diagram with a lot of pieces of the Biot-Savart Law unpacked. We show an example $\text{d}\vec{l}$, and a separation vector  $\vec{r}$ . Notice that $\text{d}\vec{l}$ is directed along the segment, in the same direction as the current. The separation vector  $\vec{r}$  points as always from source to observation.
+We can find the magnetic field from the UFO based on the compass measurement. Below we show the geometric argument.
+[{{ 184_notes:9_b_field_trig.png?400 |Tangent = Opposite over Adjacent}}]
+We know that the UFO is closest when it is directly overhead, which is also when it will have the largest contribution to the magnetic field at our observation point, resulting in the largest displacement of the compass needle. On our graph, this would be  $20^\circ$ . We can also reason that the magnetic field of the UFO at our observation point will point in the  $+x$ -direction, due to the [[184_notes:rhr|Right Hand Rule]], assuming the charge of the UFO is positive. First, you would point your fingers in the direction of the UFO's velocity then curl them toward the observation location (down) and your thumb would point out towards you or in the $+x $-direction. (If the charge of the UFO were negative, we would expect the B-field to be directed in the$ -x$-direction. This is not the case, so we can conclude that the UFO is positively charged.) This means that we can solve for the magnetic field from the UFO since we know Earth's magnetic field and the angle:
+$$\vec{B}_{\text{UFO}} = B_{\text{earth}} \tan \theta \hat{x}$$
+Now we only have to relate the B-field from the UFO to the charge, which we'll do with the Biot-Savart law we just learned about. When the UFO is directly overhead, it should not be hard to see (pictured below) that the separation vector is $\vec{r}=-h\hat{z} $and the velocity of the UFO is$ \vec{v}=-v\hat{y}$. Their cross product is then $$\vec{v} \times \vec{r} = hv \hat{x}$$
+[{{ 184_notes:9_ufo_separation.png?400 |UFO Directly Overhead}}]
+Then plugging these into the Biot-Savart law, we can find another expression for the B-field.
+$$\vec{B}_{\text{UFO}} = \frac{\mu_0}{4 \pi}\frac{q \vec{v}\times \vec{r}}{r^3}$$
+ $\vec{B}_{\text{UFO}} = \frac{\mu_0}{4 \pi}\frac{qhv}{h^3}\hat{x}$
+At this point, we have two ways to express the magnetic field of the UFO, which should help us find an expression for  $q$ , the charge of the UFO. Setting these two expressions equal to one another, we find:
+ $B_{\text{earth}} \tan \theta \hat{x} = \frac{\mu_0}{4 \pi}\frac{qhv}{h^3}\hat{x}$
+Solving for  $q$ , we have
-{{ 184_notes:9_current_segment.png?400 |Segment of Current}}
+$$q=\frac{4 \pi h^2 B_{\text{earth}} \tan \theta}{\mu_0 v} = 131 \text{ C}$$
-For now, we write  $\text{d}\vec{l} = \langle \text{d}x, \text{d}y, 0 \rangle$
+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.
-and  $\vec{r} = \vec{r}_{obs} - \vec{r}_{source} = 0 - \langle x, y, 0 \rangle = \langle -x, -y, 0 \rangle$
-Notice that we can rewrite  $y$  as  $y=-L-x$ . This is a little tricky to arrive at, but is necessary to figure out unless you rotate your coordinate axes, which would be an alternative solution to this example. If finding  $y$  is troublesome, it may be helpful to rotate. We can take the derivative of both sides to find  $\text{d}y=-\text{d}x$ . We can now plug in to express  $\text{d}\vec{l}$  and  $\vec{r}$  in terms of  $x$  and  $\text{d}x$ :
-$$\text{d}\vec{l} = \langle \text{d}x, -\text{d}x, 0 \rangle$$
- $\vec{r} = \langle -x, L+x, 0 \rangle$
-Now, a couple other quantities that we see will be useful:
- $\text{d}\vec{l} \times \vec{r} = \langle 0, 0, \text{d}x(L+x) - (-\text{d}x)(-x) \rangle = \langle 0, 0, L\text{d}x \rangle = L\text{d}x \hat{z}$
- $r^3 = (x^2 + (L+x)^2)^{3/2}$
-The last thing we need is the bounds on our integral. Our variable of integration is  $x$ , since we chose to express everything in terms of  $x$  and  $\text{d}x$ . Our segment begins at  $x=-L$ , and ends at  $x=0$ , so these will be the limits on our integral. Below, we write the integral all set up, and then we evaluate using some assistance some [[https://www.wolframalpha.com/input/?i=integral+from+-L+to+0+of+L%2F(x%5E2%2B(L%2Bx)%5E2)%5E(3%2F2)+dx|Wolfram Alpha]].
-\begin{align*}
-\vec{B} &= \int \frac{\mu_0}{4 \pi}\frac{I \cdot d\vec{l}\times \vec{r}}{r^3} \\
-        &= \int_{-L}^0 \frac{\mu_0}{4 \pi}\frac{IL\text{d}x}{(x^2 + (L+x)^2)^{3/2}}\hat{z} \\
-        &= \frac{\mu_0}{2 \pi}\frac{I}{L}\hat{z}
-\end{align*}