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--- 184_notes:examples:week12_flux_examples [2017/11/08 14:26] – created tallpaul
+++ 184_notes:examples:week12_flux_examples [2018/08/09 18:08] (current) – curdemma
@@ Line 1: / Line 1: @@
+[[184_notes:b_flux|Return to Changing Magnetic Flux notes]]
 ===== Review of Flux through a Loop =====
 Suppose you have a magnetic field  $\vec{B} = 0.6 \text{ mT } \hat{x}$ . Three identical square loops with side lengths  $L = 0.5 \text{ m}$  are situated as shown below. The perspective shows a side view of the square loops, so they appear very thin even though they are squares when viewed face on.
-{{ 184_notes:12_three_loops.png?400 |Square Loops in the B-field}}
+[{{ 184_notes:12_three_loops.png?600 |Square Loops in the B-field}}]
 ===Facts===
@@ Line 19: / Line 21: @@
   * We represent magnetic flux through an area as
  $\Phi_B = \int \vec{B} \bullet \text{d}\vec{A}$
-  * We represent the situation with the given representation in the example statement above.
+  * We represent the situation with the given representation in the example statement above. Below, we also show a side and front view of the first loop for clarity.
+[{{ 184_notes:12_first_loop.png?500 |First Loop}}]
 ====Solution====
-In order to break down our approach into manageable chunks, we split up the loop into its four sides, and proceed. It is easy to find the magnitude of the force on each side, since  $\theta$  for each side is just the angle between the magnetic field and the directed current.
+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 either), then we can simplify the dot product:
+$$\vec{B} \bullet \text{d}\vec{A} = B\text{d}A\cos\theta$$
-{{ 184_notes:12_force_theta.png?550 |Theta for Each Side}}
+Since  $B$  and  $\theta$  do not change for different little pieces ($\text{d}A$) of the area, we can pull them outside the integral:
-This gives the following magnitudes:
+ $\int B\text{d}A\cos\theta =B\cos\theta \int \text{d}A = BA\cos\theta$
+Area for a square is just  $A = L^2$ , and  $\theta$  is different for each loop:
 \[
-  \left| \vec{F} \right| = \begin{cases}
+  \Phi_B = \begin{cases}
-                             IBL\sin \pi = 0 & \text{top} \\
+             BL^2\cos 0 = 1.5 \cdot 10^{-4} \text{ Tm}^2 & \text{Loop 1} \\
-                             IBL\sin 0 = 0 & \text{bottom} \\
+             BL^2\cos 90^\text{o} = 0 & \text{Loop 2} \\
-                             IBL\sin \frac{\pi}{2} = IBL & \text{left} \\
+             BL^2\cos 42^\text{o} = 1.1  \cdot 10^{-4} \text{ Tm}^2 & \text{Loop 3}
-                             IBL\sin \frac{\pi}{2} = IBL & \text{right} \\
+           \end{cases}
-                           \end{cases}
 \]
-It remains to find the direction of the force, for which we will use the [[184_notes:rhr|Right Hand Rule]]. You should be able to convince yourself based on the coordinates we have chosen that the force on the left side is in the  $+\hat{z}$  direction, and the force on the right side is in the  $-\hat{z}$  direction.
+Notice that we could've given answers for Loops 1 and 2 pretty quickly, since they are parallel and perpendicular to the magnetic field, respectively, which both simplify the flux calculation greatly.
-This means that the net force on the loop is  $0$ , the loop's center of mass won't move! However, the opposing forces on opposite sides will cause the loop to spin -- there is a torque! The calculation for the torque is shown below, with a diagram included to show visually what happens.
-{{ 184_notes:12_loop_torque.png?400 |The Loop Rotates}}
-The calculation is here:
- $\vec{\tau} = \vec{r} \times \vec{F} = \vec{r}_\text{left} \times \vec{F}_\text{left} + \vec{r}_\text{right} \times \vec{F}_\text{right} = \left(-\frac{L}{2} \hat{x}\right) \times \left(IBL \hat{z}\right) + \left(\frac{L}{2} \hat{x}\right) \times \left(-IBL \hat{z}\right) = IBL^2 \hat{y}$