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--- 184_notes:examples:week12_flux_examples [2017/11/08 14:38] – [Solution] tallpaul
+++ 184_notes:examples:week12_flux_examples [2017/11/12 21:11] – [Review of Flux through a Loop] tallpaul
@@ Line 2: / Line 2: @@
 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 19: @@
   * 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====
 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:
@@ Line 34: / Line 35: @@
   \Phi_B = \begin{cases}
              BL^2\cos 0 = 1.5 \cdot 10^{-4} \text{ Tm}^2 & \text{Loop 1} \\
-             BL^2\cos 90^\text{o} = 0 \text{ Tm}^2 & \text{Loop 2} \\
+             BL^2\cos 90^\text{o} = 0 & \text{Loop 2} \\
              BL^2\cos 42^\text{o} = 1.1  \cdot 10^{-4} \text{ Tm}^2 & \text{Loop 3}
            \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:
-$$\vec{\tau} = \vec{r} \times \vec{F} = \vec{r}_\text{left} \times \vec{F}_\text{left} + \vec{r}_\text{right} \times \vec{F}_\text{right} = \left(-\frac{L}{2} \hat{x}\right) \times \left(IBL \hat{z}\right) + \left(\frac{L}{2} \hat{x}\right) \times \left(-IBL \hat{z}\right) = IBL^2 \hat{y}$$