184_notes:examples:week12_force_between_wires

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184_notes:examples:week12_force_between_wires [2018/07/19 13:31] – curdemma184_notes:examples:week12_force_between_wires [2018/07/19 13:32] – curdemma
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   * We represent the situation with diagram below.   * 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==== ====Solution====
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 If we wanted to write the force per length (rather than the total force), we would simply divide by L on both sides: If we wanted to write the force per length (rather than the total force), we would simply divide by L on both sides:
 $$\frac{\vec{F}_{1 \rightarrow 2}}{L}  = \frac{\mu_0 I_1 I_2}{2 \pi R} \hat{x}$$ $$\frac{\vec{F}_{1 \rightarrow 2}}{L}  = \frac{\mu_0 I_1 I_2}{2 \pi R} \hat{x}$$
-{{ 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. 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.
  • 184_notes/examples/week12_force_between_wires.txt
  • Last modified: 2021/07/13 12:16
  • by schram45