184_notes:examples:week9_current_segment

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184_notes:examples:week9_current_segment [2017/10/20 01:12] – [Solution] tallpaul184_notes:examples:week9_current_segment [2017/10/20 02:13] (current) – [Magnetic Field from a Current Segment] tallpaul
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 You may have read about how to find the [[184_notes:b_current#Magnetic_Field_from_a_Very_Long_Wire|magnetic field from a very long wire of current]]. Now, what is the magnetic field from a single segment? Suppose we have the configuration shown below. Your observation point is at the origin, and the segment of current I runs in a straight line from ⟨−L,0,0⟩ to ⟨0,−L,0⟩. You may have read about how to find the [[184_notes:b_current#Magnetic_Field_from_a_Very_Long_Wire|magnetic field from a very long wire of current]]. Now, what is the magnetic field from a single segment? Suppose we have the configuration shown below. Your observation point is at the origin, and the segment of current I runs in a straight line from ⟨−L,0,0⟩ to ⟨0,−L,0⟩.
  
-{{ 184_notes:9_current_segment.png?400 |Segment of Current}}+{{ 184_notes:9_current_segment_bare.png?200 |Segment of Current}}
  
 ===Facts=== ===Facts===
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 Below, we show a diagram with a lot of pieces of the Biot-Savart Law unpacked. We show an example d→l, and a separation vector →r. Notice that d→l is directed along the segment, in the same direction as the current. The separation vector →r points as always from source to observation. Below, we show a diagram with a lot of pieces of the Biot-Savart Law unpacked. We show an example d→l, and a separation vector →r. Notice that d→l is directed along the segment, in the same direction as the current. The separation vector →r points as always from source to observation.
  
-{{picture}}+{{ 184_notes:9_current_segment.png?400 |Segment of Current}}
  
 For now, we write d→l=⟨dx,dy,0⟩ For now, we write d→l=⟨dx,dy,0⟩
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 d→l×→r=⟨0,0,dx(L+x)−(−dx)(−x)⟩=⟨0,0,Ldx⟩=Ldxˆz d→l×→r=⟨0,0,dx(L+x)−(−dx)(−x)⟩=⟨0,0,Ldx⟩=Ldxˆz
 r3=(x2+(L+x)2)3/2 r3=(x2+(L+x)2)3/2
-The last thing we is the bounds on our integral. Our variable of integration is x, since we chose to express everything in terms of x and dx. Our segment begins at x=−L, and ends at x=0, so these will be the limits on our integral. Below, we write the integral all set up, and then we evaluate using some assistance some Wolfram Alpha.+The last thing we need is the bounds on our integral. Our variable of integration is x, since we chose to express everything in terms of x and dx. Our segment begins at x=−L, and ends at x=0, so these will be the limits on our integral. Below, we write the integral all set up, and then we evaluate using some assistance some [[https://www.wolframalpha.com/input/?i=integral+from+-L+to+0+of+L%2F(x%5E2%2B(L%2Bx)%5E2)%5E(3%2F2)+dx|Wolfram Alpha]].
 \begin{align*} \begin{align*}
 \vec{B} &= \int \frac{\mu_0}{4 \pi}\frac{I \cdot d\vec{l}\times \vec{r}}{r^3} \\ \vec{B} &= \int \frac{\mu_0}{4 \pi}\frac{I \cdot d\vec{l}\times \vec{r}}{r^3} \\
         &= \int_{-L}^0 \frac{\mu_0}{4 \pi}\frac{IL\text{d}x}{(x^2 + (L+x)^2)^{3/2}}\hat{z} \\         &= \int_{-L}^0 \frac{\mu_0}{4 \pi}\frac{IL\text{d}x}{(x^2 + (L+x)^2)^{3/2}}\hat{z} \\
-        &= +        &= \frac{\mu_0}{2 \pi}\frac{I}{L}\hat{z}
 \end{align*} \end{align*}
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  • Last modified: 2017/10/20 01:12
  • by tallpaul