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 [2017/11/02 21:01] – dmcpadden184_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 $q>0$ 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?
Line 22: Line 24:
   * We represent the two situations below.   * We represent the two situations below.
  
-{{ 184_notes:10_circular_setup.png?600 |Moving Charge in a Magnetic Field}}+[{{ 184_notes:10_circular_setup.png?600 |Moving Charge in a Magnetic Field}}]
  
 ====Solution==== ====Solution====
Line 29: 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:
 $$\vec{F}= q (v\hat{x}) \times (-B\hat{z}) = qvB\hat{y}$$ $$\vec{F}= q (v\hat{x}) \times (-B\hat{z}) = qvB\hat{y}$$
  
-{{ 184_notes:10_circular_force.png?600 |Force on the Moving Charge}}+[{{ 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 50: Line 52:
 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:10_circular_motion.png?700 |Motion of the Moving Charge}}+[{{ 184_notes:10_circular_motion.png?700 |Motion of the Moving Charge}}]
  • 184_notes/examples/week10_radius_motion_b_field.1509656496.txt.gz
  • Last modified: 2017/11/02 21:01
  • by dmcpadden