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184_notes:examples:week10_radius_motion_b_field [2017/10/31 13:21] – tallpaul | 184_notes:examples:week10_radius_motion_b_field [2018/07/03 14:01] (current) – curdemma | ||
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=====Radius of Circular Motion in a Magnetic Field===== | =====Radius of Circular Motion in a Magnetic Field===== | ||
Suppose you have a moving charge $q>0$ in a magnetic field $\vec{B} = -B \hat{z}$. The charge has a velocity of $\vec{v} = v\hat{x}$, and a mass $m$. What does the motion of the charge look like? What if the charge enters the field from a region with $0$ magnetic field? | Suppose you have a moving charge $q>0$ in a magnetic field $\vec{B} = -B \hat{z}$. The charge has a velocity of $\vec{v} = v\hat{x}$, and a mass $m$. What does the motion of the charge look like? What if the charge enters the field from a region with $0$ magnetic field? | ||
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* The field is constant. | * The field is constant. | ||
* In the case where the particle comes from outside, the field is a step function -- it goes immediately from $0$ to $B$, and we can draw a boundary. | * In the case where the particle comes from outside, the field is a step function -- it goes immediately from $0$ to $B$, and we can draw a boundary. | ||
+ | * The speed of the particle is constant (no other forces acting on the particle) | ||
===Representations=== | ===Representations=== | ||
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* We represent the two situations below. | * We represent the two situations below. | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
====Solution==== | ====Solution==== | ||
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We can recall an [[184_notes: | We can recall an [[184_notes: | ||
$$\vec{F}= q \vec{v} \times \vec{B}$$ | $$\vec{F}= q \vec{v} \times \vec{B}$$ | ||
- | So when we use the Right Hand Rule, we point our fingers in the direction of $\vec{v}$, which is $\hat{x}$. When we curl our fingers towards $\vec{B}$, which is directed towards $-\hat{z}$, we find that our thumb end up pointing in the $\hat{y}$ direction. Since $q$ is positive, $\hat{y}$ will be the direction of the force. The math should yield: | + | So when we use the Right Hand Rule, we point our fingers in the direction of $\vec{v}$, which is $\hat{x}$. When we curl our fingers towards $\vec{B}$, which is directed towards $-\hat{z}$, we find that our thumb ends up pointing in the $\hat{y}$ direction. Since $q$ is positive, $\hat{y}$ will be the direction of the force. The math should yield: |
$$\vec{F}= q (v\hat{x}) \times (-B\hat{z}) = qvB\hat{y}$$ | $$\vec{F}= q (v\hat{x}) \times (-B\hat{z}) = qvB\hat{y}$$ | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
So, the force on the charge is at first perpendicular to its motion. This is pictured above. You can imagine that as the charge' | So, the force on the charge is at first perpendicular to its motion. This is pictured above. You can imagine that as the charge' | ||
- | Finding the radius of this circular motion requires recalling that circular motion is dictated by a [[https:// | + | Finding the radius of this circular motion requires recalling that circular motion |
\begin{align*} | \begin{align*} | ||
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So now we know the radius. You can imagine that if the particle entered from outside the region of magnetic field, it would take the path of a semicircle before exiting the field in the opposite direction. Below, we show what the motion of the particle would be in each situation. | So now we know the radius. You can imagine that if the particle entered from outside the region of magnetic field, it would take the path of a semicircle before exiting the field in the opposite direction. Below, we show what the motion of the particle would be in each situation. | ||
- | {{ 184_notes: | + | [{{ 184_notes: |