Chapter 20, Problem 86GP

To determine

Current in the wires.

Answer to Problem 86GP

Current in the wires is $8.4\text{ A}$

Explanation of Solution

Given:

Diameter of two long straight aluminum wires is $0.50\text{ mm}$ .

Length of the string is $0.50\text{ m}$ and the angle of suspension with the vertical is $3°$

Formula used:

Magnetic force is calculated as

  $F=\frac{{\mu }_0{I}_1{I}_2}{2d_2}{l}_2$

Calculation:

Forces experienced by each wire is shown as below,

  

Since the net force in both horizontal and vertical direction remains constant. Thus,

  $\begin{array}{l}{F}_B=F_T\sin \theta \\ F_T\cos \theta =mg\\ F_T=\frac{mg}{\cos \theta }\\ F_B=\frac{mg}{\cos \theta }\sin \theta =mg\tan \theta ={\rho }_{Al}\pi r_{Al}^{2}Lg\tan \theta \end{array}$

The magnetic force is also equal to

  $\begin{array}{c}\frac{{\mu }_0{I}^2}{2S}L={\rho }_{Al}\pi r_{Al}^{2}Lg\tan \theta \\ I=\sqrt{\frac{2S{\rho }_{Al}\pi r_{Al}^{2}Lg\tan \theta }{{\mu }_{0}L}}\\ I=\sqrt{\frac{2\pi [2(0.50\text{ m})\sin 3°](2700{\text{ kg/m}}^3)\pi {(2.5\times {10}^{-4}\text{ m})}^2(9.8{\text{ m/s}}^2)\tan 3°}{(4\pi \times {10}^{-7}\text{ T}\cdot \text{m/A})}}\\ \text{=8}\text{.4 A}\end{array}$

Conclusion:

Thus, the current in the wires is $8.4\text{ A}$