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--- 184_notes:examples:week10_force_on_charge [2017/10/29 22:11] – [Solution] 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 39: / Line 40: @@
 $$\vec{F}_B = q \vec{v} \times \vec{B} = 1.5 \text{ mC } \cdot 4\cdot 10^{-3} \text{ T} \cdot \text{m/s } \hat{z} = 6 \mu\text{N}$$
-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 to the magnetic field, the cross product evaluates to 0. See below for both calculations.
+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*}