Chapter 20, Problem 29P

To determine

The magnitude and direction of the current flowing in the second wire.

Answer to Problem 29P

The current flowing in the upward direction in the second wire due to electric field of the first one is 13 A.

Explanation of Solution

Given:

The magnitude of the current flowing in the first wire is 24 A in the upward direction exerting an attractive force per unit length of $8.8\times {10}^{-4} \text{N/m}$ on the second parallel wire. The distance between both wires is 7 cm.

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:

Current flowing in the second wire due to the force experienced is

  $\begin{array}{l}{F}_2=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d}{l}_2\\ I_2=\frac{2\pi }{{\mu }_0}\frac{F_2}{l_2}\frac{d}{I_1}\\ I_2=\frac{\text{2π}}{{\text{4π×10}}^{\text{-7}}\text{T}\text{.m/A}}\text{(8}{\text{.8×10}}^{\text{-4}}\text{N/m)}\frac{{\text{7×10}}^{\text{-2}}\text{m}}{\text{24A}}\\ I_2=13\text{A}\end{array}$

Conclusion:

The current flowing in the upward direction in the second wire due to electric field of the first one is 13 A.