Differences

This shows you the differences between two versions of the page.

--- 184_notes:examples:week9_detecting_b [2017/10/24 00:39] – dmcpadden
+++ 184_notes:examples:week9_detecting_b [2021/07/05 21:58] (current) – [Solution] schram45
@@ Line 1: / Line 1: @@
+[[184_notes:moving_q|Return to Moving Charges make Magnetic Fields notes]]
 =====Magnetic Field near a Moving Charge=====
 You are a collector of magnetic field detectors. A fellow detector collector is trying to trim down her collection, and so it's your job to see if an old detector is still working properly, in which case it's yours! Today, you are a magnetic field detector collector, inspector, and hopefully a selector. First, you run a test in which a charged particle ($q = 15 \text{ nC}$) is sent through the detector, and you look at the detector's readings to see if the detector seems to be working properly. The particle is travelling at $2 \text{ m/s}$ and you need to predict the magnetic field's magnitude and direction at the points shown below. Be careful! -- You'll be using vectors.
-{{ 184_notes:9_detector_setup.png?400 |Detector Setup}}
+[{{ 184_notes:9_detector_setup.png?400 |Detector Setup}}]
 ===Facts===
@@ Line 16: / Line 18: @@
 ===Approximations & Assumptions===
-  * The particle can be treated as a point particle..
+  * We are only interested in the $B$-field at this specific moment in time: As the particle moves some its parameters may change (i.e. velocity, charge...). This assumption gives use fixed variables to work with at this snapshot in time, simplifying down the complexity of the model.
-  * We are only interested in the $B$-field at this specific moment in time.
 ===Representations===
   * We represent the Biot-Savart Law for 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 must approximate the particle as a point particle in order to use the magnetic field equation above. Since the problem doesn't say anything otherwise and it is a necessary approximation to solve the problem, this is an approximation we will make.
+</WRAP>
   * We represent the situation with diagram given above.
@@ Line 32: / Line 37: @@
 For Location 2, the $\vec{r}_2$ is perpendicular to the velocity vector $\vec{v}$ (the angle between them is 90 degrees). When we take this cross product, then:
-$$\vec{v}\times \vec{r}_2 = |\vec{v}||\vec{r}_2|sin(90)=2*0.5*1= 1 \text{ m}^2\text{s}^{-1}$$
+$$\vec{v}\times \vec{r}_2 = |\vec{v}||\vec{r}_2|sin(90)=2*0.5*1= 1 \text{ m}^2\text{s}^{-1} \text{ } \hat{z}$$
-Now we just need to figure out the direction using the right hand rule.
+Note that the $\hat{z}$ comes from using the right hand rule (or rather from taking the cross product of a vector in the $\hat{x}$ and a vector in the $\hat{y}$). If you point your fingers in the direction of the velocity and curl them toward location 2 your thumb would point out of the page (or in the $+\hat{x}$ direction).
-\begin{align*}
+Finally, for Location 3, we can think about $\vec{r}_3$ in terms of it's components - the part that is parallel to the velocity (which would be the same as $\vec{r}_1$) and the part that is perpendicular to the velocity (which would be the same as $\vec{r}_2$). We just learned if the vectors are parallel, the cross product will give zero - so the horizontal component of $\vec{r}_3$ will not contribute to the magnetic field at Location 3. The only part that will contribute will be the vertical component (which is the same as what we just calculated for magnetic field at Location 2). So, we know already that this cross product will give:
- \\
-\vec{v}\times \vec{r}_2 &= 1 \text{ m}^2\text{s}^{-1} \hat{z} \\
+$$\vec{v}\times \vec{r}_3 = 1 \text{ m}^2\text{s}^{-1} \hat{z}$$
-\vec{v}\times \vec{r}_3 &= 1 \text{ m}^2\text{s}^{-1} \hat{z}
-\end{align*}
+Next, we find the magnitudes of $r^3$, since that is another quantity we need to know in the Biot-Savart Law. Note that even though the cross products were the same for Locations 2 and 3, the separation distances are not! Location 3 is still further away from q when compared to Location 2.
- The other two cross-products are the same because locations 2 and 3 are equidistant from this path. Though they are difference distances from the particle itself, the $\hat{x}$ portion of location 3's separation vector does not contribute at all to the cross-product.
-Next, we find the magnitudes of $r^3$, since that is another quantity we need to know in the Biot-Savart Law.
 \begin{align*}
 {r_2}^3 &= 0.125 \text{ m}^3 \\
@@ Line 48: / Line 51: @@
 \end{align*}
-We don't including location 1 above since we already know the magnetic field is 0 at that location! Below, we give the magnetic field at all three locations.
+We don't including location 1 above since we already know the magnetic field is 0 at that location! Below, we give the magnetic field at all three locations by plugging what we found in the Biot-Savart equation from the Representations list above.
 \begin{align*}
@@ Line 55: / Line 58: @@
 \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*}
+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).