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--- 184_notes:examples:week10_helix [2017/10/31 14:22] – [Solution] tallpaul
+++ 184_notes:examples:week10_helix [2021/07/07 15:42] (current) – schram45
@@ Line 1: / Line 1: @@
+[[184_notes:q_path|Return to Path of a Charge through a Magnetic Field notes]]
 =====Helical Motion in a Magnetic Field=====
-Suppose you have a moving charge $q=16 \text{ mC} $in a magnetic field$ \vec{B} = 15 \text{ mT } \hat{y} $. The charge has a velocity of$ \vec{v} = (3\hat{x} + 2\hat{y}) \text{ m/s} $, and a mass of$ m = 1 \text{ g}$. What does the motion of the charge look like?
+Suppose you have a moving charge $q=20 \text{ mC} $in a magnetic field$ \vec{B} = 15 \text{ mT } \hat{y} $. The charge has a velocity of$ \vec{v} = (3\hat{x} + 2\hat{y}) \text{ m/s} $, and a mass of$ m = 1 \text{ g}$. What does the motion of the charge look like?
 ===Facts===
-  * There is a charge $q = 16 \text{ mC}$.
+  * There is a charge $q = 20 \text{ mC}$.
   * The charge has velocity  $\vec{v} = (3\hat{x} + 2\hat{y}) \text{ m/s}$ .
   * The charge has a mass  $m = 1 \text{ g}$ .
@@ Line 13: / Line 15: @@
 ===Approximations & Assumptions===
   * The B-field is constant.
+  * Charge and Mass of the particle are constant.
 ===Representations===
@@ Line 19: / Line 22: @@
   * We represent the situation below.
-{{ 184_notes:10_helical_setup.png?400 |Moving Charge in a Magnetic Field}}
+[{{ 184_notes:10_helical_setup.png?250 |Moving Charge in a Magnetic Field}}]
 ====Solution====
-We can recall an [[184_notes:examples:week10_radius_motion_b_field|earlier example]] where we had to find the motion of a charge in a magnetic field. You should be able to convince yourself based on that example that if our particle were moving in the  $\hat{x}$  direction, then the  $\hat{y}$ -directed magnetic field would cause circular motion in the  $xz$ -plane. The radius of this motion would be, as before,  $r = \frac{mv}{qB}$
+We can recall an [[184_notes:examples:week10_radius_motion_b_field|earlier example]] where we had to find the motion of a charge in a magnetic field. Based on that example, if our particle were moving in the  $\hat{x}$  direction, then the  $\hat{y}$ -directed magnetic field would cause circular motion in the  $xz$ -plane. The radius of this motion would be, as before,  $r = \frac{mv}{qB}$
-However, in this example the motion of the particle is a little different. There is a  $\hat{y}$  component to the velocity, which is parallel to the magnetic field -- to be clear, the velocity as a whole is not parallel to the magnetic field, but the existence of the  $\hat{y}$  component guarantees that they are not perpendicular. To see why we care about this nuance, consider this: Since the magnetic field is directed completely in the  $\hat{y}$ -direction, we know that  $\vec{v}\times\vec{B}$  will always be perpendicular to the  $\hat{y}$ -direction. In connection with this, since  $\vec{F}= q \vec{v} \times \vec{B}$ , we know that the  $\hat{y}$  component is zero,  $F_y = 0$ . Since the particle never experiences a force in the  $\hat{y}$ -direction, the  $\hat{y}$  component of the velocity will never change.
+However, in this example the motion of the particle is a little different. There is a  $\hat{y}$  component to the velocity, which is parallel to the magnetic field -- to be clear, the velocity as a whole is not parallel to the magnetic field, but the existence of the  $\hat{y}$  component guarantees that they are not perpendicular. Because the magnetic field is directed completely in the  $\hat{y}$ -direction, we know that  $\vec{v}\times\vec{B}$  will always be perpendicular to the  $\hat{y}$ -direction. In connection with this, since  $\vec{F}= q \vec{v} \times \vec{B}$ , this means that the  $\hat{y}$  component of the force is zero,  $F_y = 0$ . Since the particle never experiences a force in the  $\hat{y}$ -direction, the  $\hat{y}$  component of the velocity will never change.
-So it seems like the motion here will not be circular. The magnetic force still plays a role, though. You should be able to convince yourself that
+So it seems like the motion here will not be circular. The magnetic force still plays a role, though because the velocity also has an x-component. From the magnetic force equation, we find that the magnetic force would be:
  $\vec{F}= q \vec{v} \times \vec{B} = = q (v_x \hat{x} + v_y \hat{y}) \times (B \hat{y}) = q v_x B \hat{z}$
+<WRAP TIP>
+===Assumptions===
+Our assumptions about the Constant B-Field and chrage are essential in simplifying down the force equation. If these assumptions were not true out force could also have some time and space dependency which would make solving for and predicting the force much more dificult.
+</WRAP>
+This force would still push the particle into a circular motion (while still not affecting the particle's motion in the y-direction.) So the particle would move in an upwards spiraling path.
-We we look at the motion of the particle from the perspective of  $+y$  going into the page, we should see a circle with radius
+So when we look at the motion of the particle from the perspective of  $+y$  going into the page, we should see a circle with radius
-$$r = \frac{mv_x}{qB} = $$
+$$r = \frac{mv_x}{qB} = 10 \text{ m}$$
+<WRAP TIP>
-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:
+===Assumption===
-$$\vec{F}= q (v\hat{x}) \times (-B\hat{z}) = qvB\hat{y}$$
+We assumed the mass was constant, and this allows for uniform circular motion in the x-z plane. If the mass were not constant then the path of the particle could look much different as the radius of curvature of the particle at any instance in time/space could change.
+</WRAP>
-{{ 184_notes:10_circular_force.png?600 |Force on the Moving Charge}}
+However, because of the constant $\hat{y}$ component of the velocity, this circle is actually a helix. Two perspectives show the motion below.
-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.
-\begin{align*}
-\left| \vec{F}_B \right| &= \left| \vec{F}_{cent} \right| \\
-qvB &= \frac{m v^2}{r} \\
-r &= \frac{mv}{qB}
-\end{align*}
-Notice that the units work out when we check:
-$$\frac{\text{kg}\cdot\text{m/s}}{\text{C}\cdot\text{T}} = \frac{\text{kg}\cdot\text{m/s}}{\text{C}\cdot\frac{\text{kg}}{\text{C}\cdot\text{s}}} = \text{m}$$
-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_helical_motion.png?600 |Motion of the Moving Charge}}]