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--- 184_notes:i_thru [2018/07/24 14:22] – [What is the current enclosed?] curdemma
+++ 184_notes:i_thru [2020/08/24 13:29] (current) – dmcpadden
@@ Line 1: / Line 1: @@
 Section 21.6 in Matter and Interactions (4th edition)
-[[184_notes:amp_law|Next Page: Putting together Ampere's Law]]
+/*[[184_notes:amp_law|Next Page: Putting together Ampere's Law]]
-[[184_notes:loop|Previous Page: Magnetic Field along a Closed Loop]]
+[[184_notes:loop|Previous Page: Magnetic Field along a Closed Loop]]*/
 ===== Current through a loop =====
-Now that we have the left bit of the equation, the next step is to talk about the right side of Ampere's law - namely the $\mu_0 I_{enc}$ bit. For an Amperian loop that outside of a single wire, this part is actually rather simple, but it starts to become more complicated when we consider loops inside the wire. As we said before, if the wire is a bit thick, we can investigate what happens to the magnetic field inside the wire as well outside. That's a tough job with the Biot-Savart law and is one of the places where Ampere's law can be super useful. On this page, we will focus on the right-hand side of Ampere's law, namely, the current through our Amperian loop.
+Now that we have the left side of the equation, the next step is to talk about the right side of Ampere's law - namely the $\mu_0 I_{enc}$ bit. For an Amperian loop that is outside of a single wire, this part is actually rather simple, but it starts to become more complicated when we consider loops inside the wire. As we said before, if the wire is a bit thick, we can investigate what happens to the magnetic field inside the wire as well outside. That's a tough job with the Biot-Savart law and is one of the places where Ampere's law can be super useful. On this page, we will focus on the right-hand side of Ampere's law, namely, the current through our Amperian loop.
 {{youtube>394V31p-mrI?large}}
@@ Line 21: / Line 21: @@
 [{{  184_notes:week10_4.png?450|Thick wire with current, $I$, enclosed by an Amperian loop}}]
-For the purposes of these notes, let's assume with have a thick wire with a total uniform current, $I_{tot}$.
+For the purposes of these notes, let's assume we have a thick wire with a total uniform current, $I_{tot}$.
 === Enclosing all the current ===
@@ Line 29: / Line 29: @@
 === Enclosing some of the current ===
-When we want to find the magnetic field inside the wire, then we need pick the radius of the loop to be smaller than the radius of the wire. In this case, //some of the current will pass through the loop but not all of the current//. We need to be able to determine what fraction of the total current will pass through our loop.
+When we want to find the magnetic field inside the wire, then we need pick the radius of the loop to be smaller than the radius of the wire. In this case, **some of the current will pass through the loop but not all of the current**. We need to be able to determine what fraction of the total current will pass through our loop.
 [{{  184_notes:week10_2.png?500|Current density ($J$)}}]
-Just like how we used charge density to help us find the $Q_{encl}$ for Gauss's Law, we will use the current density to help us find the $I_{encl}$ (or the fraction of the current that passes through the Amperian loop). If we //__assume a constant, uniform current__//, then the current density (represented by a "J") is given by:
+Just like how we used [[184_notes:dq|charge density when talking about the charge from a line]], we will use the current density to help us find the $I_{encl}$ (or the fraction of the current that passes through the Amperian loop). If we //__assume a constant, uniform current__//, then the current density (represented by a "J") is given by:
 $$J=\frac{I_{tot}}{A_{tot}}$$
 where $I_{tot}$ is the total current going through the wire and $A_{tot}$ is the total cross-sectional area of the wire. The units of current density would then be $A/m^2$. Technically, current density is a vector ($\vec{J}$) that points in the direction that the charges are moving, but for our purposes, we will only be using the magnitude J.
@@ Line 39: / Line 39: @@
 Once we know the current density, we can use that to find the total enclosed current ($I_{enc}$). If we multiply the current density by the enclosed area (the area of the Amperian loop you chose), then that will give use the fraction of the current that passes through the loop. Mathematically, this would be:
-$$I_{enc} = J A_{enc} = I_{tot} \dfrac{A_{enc}}{A_{tot}}$$
+$$I_{enc} = J A_{enc} = \dfrac{I_{tot}}{A_{tot}} A_{enc}$$
 [{{  184_notes:week10_5.png?450|Current density when the amperian loop encloses an area smaller than the area of the wire}}]
-where again $A_{tot}$ is the total cross-sectional area of the wire, $A_{enc}$ is the cross-sectional area enclosed by the loop, and $I_{tot}$ is the total current in the wire. This is very similar to how you found [[184_notes:q_enc|$Q_{encl}$ with Gauss's Law]] for Electric Fields.
+where again $A_{tot}$ is the total cross-sectional area of the wire, $A_{enc}$ is the cross-sectional area enclosed by the loop, and $I_{tot}$ is the total current in the wire. This is very similar to how you will find [[184_notes:q_enc|$Q_{encl}$ with Gauss's Law]] for Electric Fields next week.