Force Between Two Parallel Currents — Question 10

Section 29-2 · Physics pages 4690–4691

Question 10 A triangular arrangement requires vector addition even when all pair forces are attractive Two fixed wires at (−a, 0) and (+a, 0) carry I toward +z. A third wire at (0, h) carries i toward +z. All are very long. Compute the force per length on the upper wire as a function of positive h. Use vacuum permeability µ0 ≃4π × 10−7 T m/A where numerical values are needed. I I i h a Schematic; labels, not drawing lengths, specify the data. (a) Find the vertical force; show that the horizontal components cancel. (b) Find the value of h that maximizes the force magnitude for fixed a, I and i, and compare the limits h approaching zero and infinity.
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Question 10 - Solution A triangular arrangement requires vector addition even when all pair forces are attractive I I i h a (a) Method and calculation r = √ a2 + h2,⃗ f± = µ0Ii 2πr2 (±aˆx −hˆy).⃗ f = − µ0Ii h π(a2 + h2) ˆy . (b) Interpretation and further calculation d dh h a2 + h2 = a2 −h2 (a2 + h2)2 = 0 ⇒h = a . |f|max = µ0Ii 2πa , |f| →0 as h →0 or h →∞. Physical check Near the base the two attractions are almost horizontal; far away both are weak. Their vertical sum therefore has an intermediate maximum.

Question 10

A triangular arrangement requires vector addition even when all pair forces are attractive

Two fixed wires at (−a,0)(-a,0) and (+a,0)(+a,0) carry I toward +z. A third wire at (0,h)(0,h) carries i toward +z. All are very long. Compute the force per length on the upper wire as a function of positive h. Use vacuum permeability μ0≃4π×10−7Tm/A\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A} where numerical values are needed.

Diagram shown in the original SVG worksheet.

  1. Find the vertical force; show that the horizontal components cancel.

  2. Find the value of h that maximizes the force magnitude for fixed a, I and i, and compare the limits h approaching zero and infinity.

Question 10 - Solution

A triangular arrangement requires vector addition even when all pair forces are attractive

Diagram shown in the original SVG worksheet.

(a) Method and calculation

r=a2+h2,f→±=μ0Ii2πr2(±ax̂−hŷ).r=\sqrt{a^2+h^2},\quad\vec f_\pm=\frac{\mu_0Ii}{2\pi r^2}(\pm a\hat x-h\hat y).f→=−μ0Iihπ(a2+h2)ŷ.\boxed{\vec f=-\frac{\mu_0Ii\,h}{\pi(a^2+h^2)}\hat y}.

(b) Interpretation and further calculation

ddhha2+h2=a2−h2(a2+h2)2=0⇒h=a.\frac d{dh}\frac h{a^2+h^2}=\frac{a^2-h^2}{(a^2+h^2)^2}=0\Rightarrow\boxed{h=a}.|f|max=μ0Ii2πa,|f|→0 as h→0 or h→∞.|f|_{\max}=\frac{\mu_0Ii}{2\pi a},\quad |f|\to0\text{ as }h\to0\text{ or }h\to\infty.

Physical check

Near the base the two attractions are almost horizontal; far away both are weak. Their vertical sum therefore has an intermediate maximum.