Differences

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--- 184_notes:examples:week12_moving_coils_flux [2017/11/10 18:31] – [Flux Through Moving Coils] tallpaul
+++ 184_notes:examples:week12_moving_coils_flux [2018/04/11 19:36] (current) – [Solution] pwirving
@@ Line 4: / Line 4: @@
 ===Facts===
   * The rings are oriented so that there is a magnetic flux through one due to to the other.
+  * There is only current in the first ring.
 ===Lacking===
@@ Line 19: / Line 20: @@
 {{ 184_notes:12_coils.png?400 |Two Coils}}
 ====Solution====
-Since the magnetic field has a uniform direction, and the area of the loop is flat (meaning $\text{d}\vec{A}$ does not change direction if we move along the area), then we can simplify the dot product:
+Before any motion happens, we can look at the rings from a side view, and represent the magnetic field as shown below. In the side view it looks like current is coming out of the page at the top of the coil and going into the page at the bottom. The direction of the magnetic field is determined using the [[184_notes:rhr|Right Hand Rule]].
-$$\int \vec{B} \bullet \text{d}\vec{A} = \int B\text{d}A\cos\theta$$
-Since $B$ and $\theta$ do not change for different little pieces ($\text{d}A$) of the area, we can pull them outside the integral:
+{{ 184_notes:12_coils_field.png?500 |Coils with Field}}
-$$\int B\text{d}A\cos\theta =B\cos\theta \int \text{d}A = BA\cos\theta$$
+When the rings begin to move towards one another, you can imagine that the second ring experiences an increase in magnetic flux, since the magnetic field is stronger closer to the ring with current. When we say the flux "increases", we imply that the flux was positive to begin with, so this means we are //__assuming that the area-vector of the second ring must point in the same direction as the magnetic field__//. In our representation, this would be to the right.
-It will be easier to concern ourselves with this value, rather than try to describe the integral calculation each time. The bar begins moving to the left and the area begins to close up, as shown below:
+Since the flux is increasing, we expect the induced current in the second ring to oppose this increase. This requires for the induced current to produce a magnetic field pointing to the left. We can use the [[184_notes:rhr|Right Hand Rule]] to determine that this means the current will go into the page at the top of the page, and out of the page at the bottom of the loop. We show the resulting induced current below.
-{{ 184_notes:12_rail_flux_decreased.png?400 |Decreased Flux}}
+{{ 184_notes:12_coils_induced_current.png?400 |Induced Current}}
-It should be easy to see the that the magnetic field and the orientation of the loop are not changing, but the area of the loop is decreasing. The flux through the loop is therefore decreasing, which indicates that a current is induced in the loop.
+In fact, this induced current and resulting magnetic field from the second ring will actually cause a changing flux through the first ring that actually increases its currently momentarily. We are able to reconcile this with conservation of energy because even though the current increases in both rings, the resulting magnetic forces between them opposes their motion.
-To determine the direction of the current, consider that when we say the flux decreases, this carries the assumption that the area vector points in the same direction as the magnetic field, which is into the page. The induced current should create a magnetic field that opposes the change in flux. The change in flux is a "decrease", so we expect the induced current to create a "positive" magnetic field, just meaning that it is pointing in the same direction as the area-vector. Since the area vector points into the page, we expect for the induced current to be clockwise. We determine this using the [[184_notes:rhr|Right Hand Rule]].
+{{youtube>gzolf12OYzE?large}}
-Alternatively, you could think of the mobile charges inside the moving bar. Since the bar is moving inside a magnetic field, mobile charges will experience an acceleration due to the magnetic force. The force is $\vec{F} = q \vec{v} \times \vec{B}$, which points down for positive charges, implying that conventional current will be clockwise, which is the same result we reached earlier. The result is shown below.
-{{ 184_notes:12_rail_current.png?300 |Induced Current}}