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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 [2017/10/29 21:54] – [Solution] tallpaul
@@ Line 33: / Line 33: @@
 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:
+$$\vec{F} = q \vec{v} \times \vec{B} = 1.5 \text{ mC } \cdot 4\cdot 10^{-3} \text{ T} \cdot \text{m/s } \hat{z} = 6 \mu\text{N}$$
+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