184_notes:examples:week10_radius_motion_b_field

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
Next revision		 Previous revision
184_notes:examples:week10_radius_motion_b_field [2017/10/31 11:34] – [Solution] tallpaul184_notes:examples:week10_radius_motion_b_field [2018/07/03 14:01] (current) – curdemma
Line 1: Line 1:
 +[[184_notes:q_path|Return to Path of a Charge through a Magnetic Field notes]]
 +
 =====Radius of Circular Motion in a Magnetic Field===== =====Radius of Circular Motion in a Magnetic Field=====
-Suppose you have a moving charge $q$ in a magnetic field $\vec{B} = -B \hat{z}$. The charge has a speed of $\vec{v} = v\hat{x}$. 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?
  
 ===Facts=== ===Facts===
   * There is a charge $q$.   * There is a charge $q$.
   * The charge has velocity $v\hat{x}$.   * The charge has velocity $v\hat{x}$.
 +  * The charge has a mass $m$.
   * The charge is in a field $\vec{B} = -B \hat{z}$.   * The charge is in a field $\vec{B} = -B \hat{z}$.
  
Line 14: Line 17:
   * 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===
Line 20: Line 24:
   * We represent the two situations below.   * We represent the two situations below.
  
-{{ 184_notes:10_circular_setup.png?500 |Moving Charge in a Magnetic Field}}+[{{ 184_notes:10_circular_setup.png?600 |Moving Charge in a Magnetic Field}}] 
 ====Solution==== ====Solution====
 Let's start with the first case, when the magnetic field exists all around the charge. Let's start with the first case, when the magnetic field exists all around the charge.
Line 26: Line 31:
 We can recall an [[184_notes:examples:week10_force_on_charge|earlier example]] where we had to find the force on a similar charge. One approach we took to find the direction of force was the [[184_notes:rhr|Right Hand Rule]]. Remember that the force here will be  We can recall an [[184_notes:examples:week10_force_on_charge|earlier example]] where we had to find the force on a similar charge. One approach we took to find the direction of force was the [[184_notes:rhr|Right Hand Rule]]. Remember that the force here will be 
 $$\vec{F}= q \vec{v} \times \vec{B}$$ $$\vec{F}= q \vec{v} \times \vec{B}$$
-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:
 $$\vec{F}= q (v\hat{x}) \times (-B\hat{z}) = qvB\hat{y}$$ $$\vec{F}= q (v\hat{x}) \times (-B\hat{z}) = qvB\hat{y}$$
  
-{{picture}}+[{{ 184_notes:10_circular_force.png?600 |Force on the Moving 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's velocity is directed a little towards the $\hat{y}$ direction, the force on it will also change a little, since the cross product that depends on velocity will change a little. In fact, if you remember from [[184_notes:q_path|the notes]], this results in circular motion if the charge is in a constant magnetic field. So, the force on the charge is at first perpendicular to its motion. This is pictured above. You can imagine that as the charge's velocity is directed a little towards the $\hat{y}$ direction, the force on it will also change a little, since the cross product that depends on velocity will change a little. In fact, if you remember from [[184_notes:q_path|the notes]], this results in circular motion if the charge is in a constant magnetic field.
  
-Finding the radius of this circular motion requires recalling that circular motion is dictated by a [[https://en.wikipedia.org/wiki/Centripetal_force|centripetal force]]. This is the same force that we computed earlier -- the magnetic force -- since this is the force that is perpendicular to the particle's motion. Below, we set the two forces equal to find the radius of the circular motion.+Finding the radius of this circular motion requires recalling that circular motion occurs when the net force on the particle is perpendicular to the velocity. When this is true, we usually call this net force the [[https://en.wikipedia.org/wiki/Centripetal_force|centripetal force]]. In this example, the centripetal force would be the same force that we computed earlier -- the magnetic force -- since this is the force that is perpendicular to the particle's motion (or only force in this case). Below, we set the two forces equal to find the radius of the circular motion.
  
 \begin{align*} \begin{align*}
Line 43: Line 48:
 Notice that the units work out when we check: Notice that the units work out when we check:
  
-$$\frac{\text{kg}\cdot\text{m/s}}{\text{C}\cdot\text{T}} = \frac{\text{kg}\cdot\text{m/s}}{\text{C}\cdot\frac{\text{kg}}{\text{C}\cdot\text{s}}}$$ +$$\frac{\text{kg}\cdot\text{m/s}}{\text{C}\cdot\text{T}} = \frac{\text{kg}\cdot\text{m/s}}{\text{C}\cdot\frac{\text{kg}}{\text{C}\cdot\text{s}}} = \text{m}$$
- +
-The trickiest part of finding magnetic force is the cross-product. You may remember from the [[184_notes:math_review#Vector_Multiplication|math review]] that there are a couple ways to do the cross product. Below, we show how to use vector components, for which it's helpful to rewrite $\vec{v}$ and $\vec{B}$ with their components. +
- +
-\begin{align*} +
-  \vec{v} &= \langle 10, 0, 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 \\ +
-                         &= \langle 0, 0, 4\cdot 10^{-3} \rangle \text{ T} \cdot \text{m/s} +
-\end{align*} +
- +
-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: +
- +
-$$\vec{F}_B = q \vec{v} \times \vec{B} = 1.5 \text{ mC } \cdot 4\cdot 10^{-3} \text{ T} \cdot \text{m/s } \hat{z} = 6 \mu\text{N}$$ +
- +
-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 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 \\ +
-                         &= \langle 0, 0, 0 \rangle +
-\end{align*} +
- +
-Or, with whole vectors:+
  
-$$\left|\vec{v} \times \vec{B} \right|= \left|\vec{v}\right| \left|\vec{B} \right| \sin\theta = (10 \text{ m/s})(4\cdot 10^{-4} \text{ T}) \sin 0 = 0$$+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.
  
-When the velocity is parallel to the magnetic field, $\vec{F}_B=0$.+[{{ 184_notes:10_circular_motion.png?700 |Motion of the Moving Charge}}]
  • 184_notes/examples/week10_radius_motion_b_field.1509449696.txt.gz
  • Last modified: 2017/10/31 11:34
  • by tallpaul