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--- 184_notes:examples:week10_force_on_charge [2017/10/29 21:42] – [Solution] 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====
 Let's start with the first case, when $\vec{v}=10 \text{ m/s } \hat{x}$.
@@ Line 33: / Line 34: @@
 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, $\vec{F}_B=0$.