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--- 184_notes:loop [2018/11/01 21:27] – dmcpadden
+++ 184_notes:loop [2020/08/24 13:28] – dmcpadden
@@ Line 1: / Line 1: @@
 Section 21.6 in Matter and Interactions (4th edition)
-[[184_notes:i_thru|Next Page: Current through a Loop]]
+/*[[184_notes:i_thru|Next Page: Current through a Loop]]
-[[184_notes:motiv_amp_law|Previous page: Motivating Ampere's Law]]
+[[184_notes:motiv_amp_law|Previous page: Motivating Ampere's Law]]*/
 ===== Magnetic field along a closed loop =====
@@ Line 55: / Line 55: @@
 This means we want to choose a closed loop that: 1) is always following the B-Field (to make $d\vec{l}$ parallel to $\vec{B}$) and 2) has a constant B-Field everywhere around the loop. So if the magnetic field points in circle around the wire, you want to pick a circular loop. If the magnetic field is constant, you want to pick a loop with at least one straight edge (like a rectangular loop).
-//It's important to note that this loop isn't real// - there is not a wire or anything physically around the wire. It's an imagined loop that helps guide our work with Ampere's Law. We refer to this imagined loop as the "Amperian Loop."
+**It's important to note that this loop isn't real** - there is not a wire or anything physically around the wire. It's an imagined loop that helps guide our work with Ampere's Law. We refer to this imagined loop as the "Amperian Loop."
 [{{  184_notes:week10_3.png?400|Amperian loop around a line of current}}]
@@ Line 74: / Line 74: @@
 \end{align*}
-where the first two steps listed come from the fact that $\vec{B}$ points in the direction of $d\vec{l}$ everywhere on the loop (so the dot product simplifies to multiplication) and the last two steps listed come from the fact that the B-field is the same magnitude along that loop (so you can pull it out of the integral). The resulting integral of $dl$ gives just $l$. Because this length comes from the integration of the $dl$, the length $l$ here represents the length of the Amperian loop. For our example, the length of the loop would just be the circumference of the loop. Since we said that the loop would have a radius R, this means $l=2\pi R$, so we get:
+where the first two steps listed come from the fact that $\vec{B}$ points in the direction of $d\vec{l}$ everywhere on the loop (so the dot product simplifies to multiplication) and the last two steps listed come from the fact that the B-field is the same magnitude along that loop (so you can pull it out of the integral). The resulting integral of $dl$ gives just $l$. Because this length comes from the integration of the $dl$, **the length $l$ here represents the length of the Amperian loop**. For our example, the length of the loop would just be the circumference of the loop. Since we said that the loop would have a radius R, this means $l=2\pi R$, so we get:
 $$\oint \vec{B} \bullet d\vec{l} = B(2 \pi R)$$
 Remember that the $B$ in this equation is the magnitude of the B-field on that loop (so this would be the magnetic field at a distance $R$ away from the straight wire).