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--- 184_notes:examples:week10_force_on_charge [2017/10/29 21:54] – [Magnetic Force on Moving Charge] tallpaul
+++ 184_notes:examples:week10_force_on_charge [2017/11/02 13:32] (current) – dmcpadden
@@ 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====
 Let's start with the first case, when  $\vec{v}=10 \text{ m/s } \hat{x}$ .
@@ Line 37: / Line 38: @@
 We get the same answer with both methods. Now, for the force calculation:
- $\vec{F} = q \vec{v} \times \vec{B} = 1.5 \text{ mC } \cdot 4\cdot 10^{-3} \text{ T} \cdot \text{m/s } \hat{z} = 6 \mu\text{N}$
+$$\vec{F}_B = q \vec{v} \times \vec{B} = 1.5 \text{ mC } \cdot 4\cdot 10^{-3} \text{ T} \cdot \text{m/s } \hat{z} = 6 \mu\text{N}$$
-Notice that the  $\sin 90^{\text{o}}$  term evaluates to  $1$ . This is the maximum value it can be, and we get this value because the velocity and the B-field were exactly perpendicular. Any other orientation would have yielded a weaker magnetic force. In fact, in the second case, when the velocity is parallel
+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,  $\vec{F}_B=0$ .