Force Between Two Parallel Currents — Question 4

Section 29-2 · Physics pages 4678–4679

Question 4 Four corner currents exert a diagonal force on each wire Four infinite parallel wires lie at the corners of a square of side a in the xy plane. All carry the same current I toward +z. Find the force per unit length on the wire at the lower-left corner, with +x rightward and +y upward. Use vacuum permeability µ0 ≃4π × 10−7 T m/A where numerical values are needed. a all currents out of page Schematic; labels, not drawing lengths, specify the data. (a) Add the force vectors from the two adjacent wires and the opposite corner. (b) Evaluate the magnitude for I=10.0 A and a=.0200 m.
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Question 4 - Solution Four corner currents exert a diagonal force on each wire a all currents out of page (a) Method and calculation C = µ0I2 2πa ,⃗ fadjacent = C(ˆx + ˆy).⃗ fdiagonal = C √ 2 ˆx + ˆy √ 2 = C 2 (ˆx + ˆy),⃗ f = 3C 2 (ˆx + ˆy). (b) Interpretation and further calculation C = 1.00 × 10−3 N/m, |f| = 3 √ 2 2 C. |f| = 2.12132 × 10−3 N/m toward the square center. Physical check The diagonal wire is farther away and its force must also be resolved into components; it does not contribute a full adjacent-wire force to either axis.

Question 4

Four corner currents exert a diagonal force on each wire

Four infinite parallel wires lie at the corners of a square of side a in the xy plane. All carry the same current I toward +z. Find the force per unit length on the wire at the lower-left corner, with +x rightward and +y upward. 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. Add the force vectors from the two adjacent wires and the opposite corner.

  2. Evaluate the magnitude for I=10.0 A and a=.0200 m.

Question 4 - Solution

Four corner currents exert a diagonal force on each wire

Diagram shown in the original SVG worksheet.

(a) Method and calculation

C=μ0I22πa,f→adjacent=C(x̂+ŷ).C=\frac{\mu_0I^2}{2\pi a},\quad\vec f_{\mathrm{adjacent}}=C(\hat x+\hat y).f→diagonal=C2x̂+ŷ2=C2(x̂+ŷ),f→=3C2(x̂+ŷ).\vec f_{\mathrm{diagonal}}=\frac C{\sqrt2}\frac{\hat x+\hat y}{\sqrt2}=\frac C2(\hat x+\hat y),\quad\vec f=\frac{3C}{2}(\hat x+\hat y).

(b) Interpretation and further calculation

C=1.00×10−3N/m,|f|=322C.C=1.00\times10^{-3}\,\mathrm{N/m},\quad|f|=\frac{3\sqrt2}{2}C.|f|=2.12132×10−3N/mtoward the square center.\boxed{|f|=2.12132\times10^{-3}\,\mathrm{N/m}}\quad\text{toward the square center}.

Physical check

The diagonal wire is farther away and its force must also be resolved into components; it does not contribute a full adjacent-wire force to either axis.