Chapter 20, Problem 39P

To determine

The force per unit length exerted on the other two wires.

Answer to Problem 39P

The force per unit length exerted on the nearest wire is $4.5\times {10}^{-2}\text{N/m}$ attractive and on the far one is $2.2\times {10}^{-2}\text{N/m}$ repelling.

Explanation of Solution

Given:

The magnitude of the current flowing in both the wires is 25 A in the upward and downward direction respectively. The distance between the third wire and the nearest one is 2.8mm.

  

Formula used:

Force exerted by a single wire on the other parallel single wire is $F_2=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d}{I}_2$

Calculation:

The force experienced in the third wire due to the nearest one is

  $\begin{array}{l}{F}_{near}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d_{near}}{l}_{near}\\ \frac{F_{near}}{l_{near}}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d_{near}}\\ \frac{F_{near}}{l_{near}}=\frac{(4\pi \times {10}^{-7})}{2\pi }\frac{{(25)}^2}{(2.8\times {10}^{-3})}\\ \frac{F_{near}}{l_{near}}=4.46\times {10}^{-2}\text{N/m}\end{array}$

And, the force experienced in the third wire due to the far one is

  $\begin{array}{l}{F}_{far}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d_{far}}{l}_{far}\\ \frac{F_{far}}{l_{far}}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d_{far}}\\ \frac{F_{far}}{l_{far}}=\frac{{\text{(4π×10}}^{\text{-7}}\text{T}\text{.m/A)}}{\text{2π}}\frac{{\text{(25A)}}^{\text{2}}}{\text{(5}{\text{.6×10}}^{\text{-3}}\text{m)}}\\ \frac{F_{far}}{l_{far}}=2.2\times {10}^{-2}\text{N/m}\end{array}$

Conclusion:

The force per unit length exerted on the nearest wire is $4.5\times {10}^{-2}\text{N/m}$ attractive and on the far one is $2.2\times {10}^{-2}\text{N/m}$ repelling.