Chapter 20, Problem 74GP

To determine

The current required to flow in the suspended copper wire.

Answer to Problem 74GP

The required current flow in the suspended copper wire 79 A.

Explanation of Solution

Given:

Three wires are at $3.8 \text{cm}$ away from each other. The direction of current in wire M is opposite to that in the wires N and P. The top wire M is of 1 mm diameter copper wire. The current flowing through the bottom two wires is 95 A each.

  

Formula used:

The magnetic force in the wire is calculated as

  $F=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}a$

Calculation:

Both N and P wires will repel the M wire since the direction of the current in the M wire is opposite to the others. The repelling forces will be along the line directly from the wires N and P away from the M wire. Only the vertical components of those forces will be counteracting gravity.

  $\begin{array}{l}mg=2F_B\cos 30°\\ {\rho }_{Cu}L\pi {\left(\frac{1}{2}{d}_{Cu}^2\right)}^{2}g=2\frac{{\mu }_0}{2\pi }\frac{I_{bottom}{I}_{Cu}}{d}L\cos 30°\\ I_{Cu}=\frac{{\rho }_{Cu}{\pi }^2{\left(\frac{1}{2}{d}_{Cu}^2\right)}^{2}dg}{{\mu }_0{I}_{bottom}}\text{=}\frac{{\text{(8900kg/m}}^{\text{3}}{\text{)π}}^{\text{2}}{{\text{(5×10}}^{\text{-4}}\text{m)}}^{\text{2}}\text{(0}\text{.038m)(9}{\text{.8m/s}}^{\text{2}}\text{)}}{{\text{(4π×10}}^{\text{-7}}{\text{T-m/A)(95A)(cos30}}^{\text{o}}\text{)}}\\ I_{Cu}=79\text{A}\end{array}$

Conclusion:

The required current flow in the suspended copper wire is 79 A.