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184_notes:examples:week10_force_on_charge [2017/10/29 20:11] – created tallpaul | 184_notes:examples:week10_force_on_charge [2017/11/02 13:32] (current) – dmcpadden | ||
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- | =====Magnetic | + | =====Magnetic |
- | You may have read about how to find the [[184_notes: | + | 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 |
- | + | ||
- | {{ 184_notes: | + | |
===Facts=== | ===Facts=== | ||
- | * The current in the segment | + | * The charge |
- | * The observation point is at the origin. | + | * There is an external magnetic field $\vec{B} = 0.4 \text{ mT } \hat{y}$. |
- | * The segment stretches from from $\langle -L, 0, 0 \rangle$ to $\langle 0, -L, 0 \rangle$. | + | * The velocity of the charge is $\vec{v} = 10 \text{ m/s } \hat{x}$ or $\vec{v} = -10 \text{ m/s } \hat{y}$. |
===Lacking=== | ===Lacking=== | ||
- | * $\vec{B}$ | + | * $\vec{F}_B$ |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * The current is steady, and the wire segment | + | * The magnetic force on the charge contains no unknown contributions. |
+ | * The charge | ||
===Representations=== | ===Representations=== | ||
- | * We represent the Biot-Savart Law for magnetic | + | * We represent the magnetic |
- | $$\vec{B}= \int \frac{\mu_0}{4 \pi}\frac{I \cdot d\vec{l}\times \vec{r}}{r^3}$$ | + | $$\vec{F}= q \vec{v} \times \vec{B}$$ |
- | * We represent the situation with diagram given above. | + | * We represent the two situations below. |
+ | {{ 184_notes: | ||
====Solution==== | ====Solution==== | ||
- | Below, we show a diagram | + | Let's start with the first case, when $\vec{v}=10 \text{ |
- | {{ 184_notes:9_current_segment.png? | + | The trickiest part of finding magnetic force is the cross-product. You may remember from the [[184_notes:math_review# |
- | For now, we write $$\text{d}\vec{l} = \langle \text{d}x, \text{d}y, 0 \rangle$$ | ||
- | and $$\vec{r} = \vec{r}_{obs} - \vec{r}_{source} = 0 - \langle x, y, 0 \rangle = \langle -x, -y, 0 \rangle$$ | ||
- | Notice that we can rewrite $y$ as $y=-L-x$. This is a little tricky to arrive at, but is necessary to figure out unless you rotate your coordinate axes, which would be an alternative solution to this example. If finding $y$ is troublesome, | ||
- | $$\text{d}\vec{l} = \langle \text{d}x, -\text{d}x, 0 \rangle$$ | ||
- | $$\vec{r} = \langle -x, L+x, 0 \rangle$$ | ||
- | Now, a couple other quantities that we see will be useful: | ||
- | $$\text{d}\vec{l} \times \vec{r} = \langle 0, 0, \text{d}x(L+x) - (-\text{d}x)(-x) \rangle = \langle 0, 0, L\text{d}x \rangle = L\text{d}x \hat{z}$$ | ||
- | $$r^3 = (x^2 + (L+x)^2)^{3/ | ||
- | The last thing we need is the bounds on our integral. Our variable of integration is $x$, since we chose to express everything in terms of $x$ and $\text{d}x$. Our segment begins at $x=-L$, and ends at $x=0$, so these will be the limits on our integral. Below, we write the integral all set up, and then we evaluate using some assistance some [[https:// | ||
\begin{align*} | \begin{align*} | ||
- | \vec{B} &= \int \frac{\mu_0}{4 \pi}\frac{I | + | |
- | & | + | |
- | &= \frac{\mu_0}{2 \pi}\frac{I}{L}\hat{z} | + | \vec{v} \times \vec{B} & |
+ | | ||
\end{align*} | \end{align*} | ||
+ | |||
+ | Alternatively, | ||
+ | |||
+ | $$\left|\vec{v} \times \vec{B} \right|= \left|\vec{v}\right| \left|\vec{B} \right| \sin\theta = (10 \text{ m/ | ||
+ | |||
+ | 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 \\ | ||
+ | & | ||
+ | \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/ | ||
+ | |||
+ | When the velocity is parallel to the magnetic field, $\vec{F}_B=0$. |