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--- 184_notes:examples:week14_changing_current_rectangle [2018/01/25 16:27] – caballero
+++ 184_notes:examples:week14_changing_current_rectangle [2021/07/13 13:26] (current) – [Solution] schram45
@@ Line 1: / Line 1: @@
-===== Changing Current Induces Voltage in Rectangular Loops =====
+[[184_notes:b_flux_t|Return to Changing Magnetic Fields with Time notes]]
+===== Changing Current Induces Voltage in Rectangular Loop =====
 Suppose you have an increasing current through a long wire,  $I(t) = I_0 \frac{t}{t_0}$ . Next to this wire, there is a rectangular loop of width  $w$  and height  $h$ . The side of the rectangle is aligned parallel to the wire so that the rectangle is a distance  $d$  from the wire, and they are both in the same plane. What is the induced voltage in the rectangle? In what direction is the induced current in the rectangle?
@@ Line 12: / Line 14: @@
 ===Approximations & Assumptions===
-  * The long wire is infinitely long and thin and straight.
+  * The long wire is infinitely long and thin and straight: With these assumptions the magnetic field produced by the wire only depends on the radial distance away from the wire and the current in it. This also allows us to use a simplified magnetic field equation from the notes.
-  * There are no external contributions to the B-field.
+  * There are no external contributions to the B-field: We are not told about any other external moving charges or currents that could also produce a magnetic field that would effect the flux through our loop, so we will assume the only contribution is from the long wire.
 ===Representations===
@@ Line 21: / Line 23: @@
   * We represent the situation with the following visual. We arbitrarily choose a direction for the current.
-{{ 184_notes:14_wire_rectangle.png?500 |Wire and Rectangle}}
+[{{ 184_notes:14_wire_rectangle.png?500 |Wire and Rectangle}}]
 ====Solution====
 In order to find the induced voltage, we will need the magnetic flux. This requires defining an area-vector and determining the magnetic field. We can use the [[184_notes:rhr|right hand rule]] to determine that the magnetic field from the wire is into the page ( $-\hat{z}$ ) near the rectangle (point your fingers in the direction of the current and curl them toward the rectangle - your thumb should point into the page). For convenience, we will also define for the area vector to be into the page. Since they both point in the same direction, the dot product simplifies:
@@ Line 33: / Line 35: @@
 In the first equality, we just changed the integral from over the "rectangle" to be over its two dimensions, "left to right", and "top to bottom". We also changed  $\text{d}A$  to be in terms of its two dimensions,  $\text{d}A = \text{d}x\text{d}y$ . A visual is shown below for clarity.
-{{ 184_notes:14_da_dx_dy.png?500 |Picture of dA Breakdown}}
+[{{ 184_notes:14_da_dx_dy.png?500 |Picture of dA Breakdown}}]
 At this point, our integral is set up enough that we can crunch through the analysis. We can pull out constants to the front and calculate to the end:
@@ Line 50: / Line 52: @@
 Notice that the induced voltage is negative. This means that the induced current produces a magnetic fields whose corresponding flux is negative flux. Since the area-vector was defined as into the page ( $-\hat{z}$ ), this means that the magnetic field produced by the induced current should be out of the page ( $+\hat{z}$ ). By the right hand rule, this means the induced currents is directed //counterclockwise//. See below for a visual.
-{{ 184_notes:14_rectangle_induced_current.png?500 |Induced Current}}
+[{{ 184_notes:14_rectangle_induced_current.png?500 |Induced Current}}]
+The induced voltage in this problem was a constant, which means the flux must be changing linearly with time. Thats the only way the derivative can be a constant. If we look at the equation for flux through a loop there are three ways it can change. The first is if the magnetic field changes, the second is if the cross product changes, and the third is if the area changes with time. None of the sides of our loop are moving, so the area is not changing with time. The relationship of our magnetic field and area vector are also not changing as our loop is not rotating and our magnetic field is always in the same direction. This leaves the magnetic field, which is changing in this example. The magnetic field in this problem changes just like the current in the long wire. Since the current in the long wire changes linearly with time, our flux must change linearly with time as well. This means our induced voltage should be a constant as our solution shows.