Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision | Last revisionBoth sides next revision | ||
184_notes:examples:week10_force_on_charge [2017/10/29 22:23] – [Solution] tallpaul | 184_notes:examples:week10_force_on_charge [2017/10/29 22:24] – [Solution] tallpaul | ||
---|---|---|---|
Line 39: | Line 39: | ||
$$\vec{F}_B = q \vec{v} \times \vec{B} = 1.5 \text{ mC } \cdot 4\cdot 10^{-3} \text{ T} \cdot \text{m/s } \hat{z} = 6 \mu\text{N}$$ | $$\vec{F}_B = q \vec{v} \times \vec{B} = 1.5 \text{ mC } \cdot 4\cdot 10^{-3} \text{ T} \cdot \text{m/s } \hat{z} = 6 \mu\text{N}$$ | ||
- | 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 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 both calculations. | + | 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 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*} | \begin{align*} |