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--- 184_notes:examples:week10_force_on_charge [2017/10/29 20:53] – [Magnetic Field from a Current Segment] tallpaul
+++ 184_notes:examples:week10_force_on_charge [2017/11/02 13:32] (current) – dmcpadden
@@ Line 1: / Line 1: @@
 =====Magnetic Force on Moving Charge=====
-Suppose you have a moving charge ( $q=1.5 \text{ mC}$ ) in a magnetic field ( $\vec{B} = 0.4 \text{ mT } \hat{y}$ ). The charge has a speed of  $10 \text{ m/s}$ . What is the magnetic force on the charge if its motion is in the  $+x$ -direction? The $+y$-direction?
+Suppose you have a moving charge ( $q=1.5 \text{ mC}$ ) in a magnetic field ( $\vec{B} = 0.4 \text{ mT } \hat{y}$ ). The charge has a speed of  $10 \text{ m/s}$ . What is the magnetic force on the charge if its motion is in the  $+x$ -direction? The $-y$-direction?
 ===Facts===
   * The charge is  $q=1.5 \text{ mC}$ .
   * There is an external magnetic field  $\vec{B} = 0.4 \text{ mT } \hat{y}$ .
-  * The velocity of the charge is  $\vec{v} = 10 \text{ m/s } \hat{x}$  or  $\vec{v} = 10 \text{ m/s } \hat{y}$ .
+  * The velocity of the charge is  $\vec{v} = 10 \text{ m/s } \hat{x}$  or $\vec{v} = -10 \text{ m/s } \hat{y}$.
 ===Lacking===
@@ Line 12: / Line 12: @@
 ===Approximations & Assumptions===
   * The magnetic force on the charge contains no unknown contributions.
+  * The charge is moving at a constant speed (no other forces acting on it)
 ===Representations===
@@ Line 18: / Line 19: @@
   * We represent the two situations below.
-{{ 184_notes:10_moving_charge.png?200 |Moving Charge in a Magnetic Field}}
+{{ 184_notes:10_moving_charge.png?500 |Moving Charge in a Magnetic Field}}
 ====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.
+Let's start with the first case, when $\vec{v}=10 \text{ m/s } \hat{x}$.
-{{ 184_notes:9_current_segment.png?400 |Segment of Current}}
+The trickiest part of finding magnetic force is the cross-product. You may remember from the [[184_notes:math_review#Vector_Multiplication|math review]] that there are a couple ways to do the cross product. Below, we show how to use vector components, for which it's helpful to rewrite $\vec{v} $and$ \vec{B}$ with their components.
-For now, we write  $\text{d}\vec{l} = \langle \text{d}x, \text{d}y, 0 \rangle$
-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} \\
+  \vec{v} &= \langle 10, 0, 0 \rangle \text{ m/s} \\
-        &= \int_{-L}^0 \frac{\mu_0}{4 \pi}\frac{IL\text{d}x}{(x^2 + (L+x)^2)^{3/2}}\hat{z} \\
+  \vec{B} &= \langle 0, 4\cdot 10^{-4}, 0 \rangle \text{ T} \\
-        &= \frac{\mu_0}{2 \pi}\frac{I}{L}\hat{z}
+  \vec{v} \times \vec{B} &= \langle v_y B_z - v_z B_y, v_z B_x - v_x B_z, v_x B_y - v_y B_x \rangle \\
+                         &= \langle 0, 0, 4\cdot 10^{-3} \rangle \text{ T} \cdot \text{m/s}
 \end{align*}
+Alternatively, we could use the whole vectors and the angle between them. We find that we obtain the same result for the cross product. One would need to use the [[184_notes:rhr|Right Hand Rule]] to find that the direction of the cross product is  $+\hat{z}$ . The magnitude is given by
+ $\left|\vec{v} \times \vec{B} \right|= \left|\vec{v}\right| \left|\vec{B} \right| \sin\theta = (10 \text{ m/s})(4\cdot 10^{-4} \text{ T}) \sin 90^{\text{o}} = 4\cdot 10^{-3} \text{ T} \cdot \text{m/s}$
+We get the same answer with both methods. Now, for the force calculation:
+ $\vec{F}_B = q \vec{v} \times \vec{B} = 1.5 \text{ mC } \cdot 4\cdot 10^{-3} \text{ T} \cdot \text{m/s } \hat{z} = 6 \mu\text{N}$
+Notice that the  $\sin 90^{\text{o}}$  is equal to  $1$ . This means that this is the maximum force that the charge can feel, and we get this value because the velocity and the B-field are exactly perpendicular. Any other orientation would have yielded a weaker magnetic force. In fact, in the second case, when the velocity is parallel to the magnetic field, the cross product evaluates to  $0$ . See below for the calculations.
+\begin{align*}
+  \vec{v} &= \langle 0, -10, 0 \rangle \text{ m/s} \\
+  \vec{B} &= \langle 0, 4\cdot 10^{-4}, 0 \rangle \text{ T} \\
+  \vec{v} \times \vec{B} &= \langle v_y B_z - v_z B_y, v_z B_x - v_x B_z, v_x B_y - v_y B_x \rangle \\
+                         &= \langle 0, 0, 0 \rangle
+\end{align*}
+Or, with whole vectors:
+ $\left|\vec{v} \times \vec{B} \right|= \left|\vec{v}\right| \left|\vec{B} \right| \sin\theta = (10 \text{ m/s})(4\cdot 10^{-4} \text{ T}) \sin 0 = 0$
+When the velocity is parallel to the magnetic field,  $\vec{F}_B=0$ .