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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 [2017/10/29 22:11] – [Solution] tallpaul
@@ Line 37: / Line 37: @@
 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 $\sin 90^{\text{o}}$ 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.
+\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, $\vec{F}_B=0$.