184_notes:examples:week12_force_between_wires [Projects & Practices in Physics]

Link to this comparison view

--- 184_notes:examples:week12_force_between_wires [2017/11/12 17:08] – dmcpadden
+++ 184_notes:examples:week12_force_between_wires [2021/07/13 12:16] (current) – schram45
@@ Line 1: / Line 1: @@
+[[184_notes:i_b_force|Return to Magnetic Force on a Current Carrying Wire notes]]
 ===== Magnetic Force between Two Current-Carrying Wires =====
 Two parallel wires have currents in opposite directions,  $I_1$  and  $I_2$ . They are situated a distance  $R$  from one another. What is the force per length  $L$  of one wire on the other?
@@ Line 10: / Line 12: @@
 ===Approximations & Assumptions===
-  * The currents are steady.
+  * The currents are steady: There are many things that could cause the current in these wires to not be steady. Since we are not told anything except the fact these currents exist, we will assume them to be steady to simplify down the model.
-  * The wires are infinitely long.
+  * The wires are infinitely long: When the wires are infinitely long the magnetic field from each wire becomes only a function of radial distance away from the wire. This also lets us use a simpler equation derived in the notes for a long wire.
-  * There are no outside forces to consider.
+  * There are no outside forces to consider: We are not told anything about external magnetic fields or the mass of the wire, so we will omit forces due to gravity, external fields, and other forces.
 ===Representations===
@@ Line 21: / Line 23: @@
   * We represent the situation with diagram below.
-{{ 184_notes:12_two_wires_representation.png?400 |Two Wires}}
+[{{ 184_notes:12_two_wires_representation.png?400 |Two Wires}}]
 ====Solution====
@@ Line 37: / Line 39: @@
 When we take the cross product of  $d\vec{l}_2$  and  $B_1$ , this gives
  $d {\vec{l}}_{2} \times {\vec{B}}_{1} = d y \hat{y} \times \frac{μ_{0} I_{1}}{2 π R} \hat{z}$  $\text{d}\vec{l}_2 \times \vec{B}_1 = \text{d}y \hat{y} \times \frac{\mu_0 I_1}{2 \pi R} \hat{z}$
-$$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \biggl(\hat{x}\times\hat{z}\biggr)$$
+$$\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \biggl(\hat{y}\times\hat{z}\biggr)$$
  $d {\vec{l}}_{2} \times {\vec{B}}_{1} = \frac{μ_{0} I_{1}}{2 π R} d y \hat{x}$  $\text{d}\vec{l}_2 \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi R}\text{d}y \hat{x}$
@@ Line 46: / Line 48: @@
 If we wanted to write the force per length (rather than the total force), we would simply divide by L on both sides:
  $\frac{{\vec{F}}_{1 \to 2}}{L} = \frac{μ_{0} I_{1} I_{2}}{2 π R} \hat{x}$  $\frac{\vec{F}_{1 \rightarrow 2}}{L}  = \frac{\mu_0 I_1 I_2}{2 \pi R} \hat{x}$
-{{ 184_notes:12_force_per_length.png?500 |Force Per Length}}
+[{{ 184_notes:12_force_per_length.png?500 |Force Per Length}}]
 If instead we wanted to find the force per length on Wire 1 from Wire 2, then we could do this whole process over again (find the magnetic field from Wire 2 at the location of Wire 1, find  $d\vec{l}_1$ , take the cross product, integrate over a segment of the wire, divide by L). We can also use what we know about forces and Newton's 3rd law, to say that the force from Wire 2 on Wire 1 should be equal and opposite to the force from Wire 1 on Wire 2.