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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 [2017/10/29 22:23] – [Solution] tallpaul | ||
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* We represent the two situations below. | * We represent the two situations below. | ||
- | {{ 184_notes: | + | {{ 184_notes: |
====Solution==== | ====Solution==== | ||
Let's start with the first case, when $\vec{v}=10 \text{ m/s } \hat{x}$. | Let's start with the first case, when $\vec{v}=10 \text{ m/s } \hat{x}$. | ||
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$$\vec{F}_B = q \vec{v} \times \vec{B} = 1.5 \text{ mC } \cdot 4\cdot 10^{-3} \text{ T} \cdot \text{m/s } \hat{z} = 6 \mu\text{N}$$ | $$\vec{F}_B = q \vec{v} \times \vec{B} = 1.5 \text{ mC } \cdot 4\cdot 10^{-3} \text{ T} \cdot \text{m/s } \hat{z} = 6 \mu\text{N}$$ | ||
- | 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 | + | 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 |
\begin{align*} | \begin{align*} |