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| 184_notes:examples:week9_current_segment [2017/10/20 01:12] – [Solution] tallpaul | 184_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⟩. |
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| {{ 184_notes:9_current_segment.png?400 |Segment of Current}} | {{ 184_notes:9_current_segment_bare.png?200 |Segment of Current}} |
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| ===Facts=== | ===Facts=== |
| 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. |
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| {{picture}} | {{ 184_notes:9_current_segment.png?400 |Segment of Current}} |
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| 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
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| r3=(x2+(L+x)2)3/2
| r3=(x2+(L+x)2)3/2
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| 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*} |