91Ó°ÊÓ

1. Current through outer wires is,$i=5.0 \text{A}$,$i_T=5.0 \text{A}=i_b=i$.
2. Current through the central wire is, $i_c=3.2 \text{A}$.
3. The length of the section of the top wire is, $L=3.0 \text{m}$.
4. The separation between any two wires is,$d=10×{10}^{-2} \text{m}$.

Parallel wires carrying current in the same direction attract each other, and the parallel wires carrying current in the opposite direction repel each other.

Formula:

The magnitude of the force on a 3.0 m section of top wire if the current in the central wire is in the positive x direction, is given by,

$F_{Tc}=\left(\frac{{\mathrm{μ}}_0{i}_c{i}_{T}L}{2\mathrm{π}d}\right)$

This is a force on T due to C. Here T and C stand for top wire and central wire.

Similarly, we can write,

$F_{Tb}=\left(\frac{{\mathrm{μ}}_0{i}_b{i}_{T}L}{2\mathrm{π}2d}\right)$

This is a force on T due to b, and b stands for a bottom wire.

Hence, the total force acting on T is,

$\begin{array}{rcl}F& =& F_{Tc}+F_{Tb}\\ & =& i_{T}L·\frac{{\mathrm{μ}}_0}{2\mathrm{π}d}\left(i_c+\frac{i_b}{2}\right)\\ F& =& \frac{2{\mathrm{μ}}_0{i}_{T}L}{4\mathrm{π}d}\left(i_c+\frac{i_b}{2}\right)\\ & & \end{array}$

Substitute all the values in the above equation.

$\begin{array}{rcl}F& =& \frac{2×\left(4\mathrm{π}×{10}^{-7} \text{T.m}/\text{A}\right)×5.0 \text{A}×3.0 \text{m}}{4\mathrm{π}×10×{10}^{-2} \text{m}}\left(3.2 \text{A}+\frac{5.0 \text{A}}{2}\right)\\ & =& 1.7×{10}^{-4} \text{N}\\ & & \end{array}$

All wires carry current in the same direction, so they attract each other. The top wire is pulled down by the other two wires.

Hence the force positive x-direction islocalid="1662852182289" $1.7×{10}^{-4} \text{N}$.

The magnitude of the force on a 3.0 m section of top wire if the current in the central wire is in the negative x direction, is given by

$F_{Tb}=\left(\frac{{\mathrm{μ}}_0{i}_b{i}_{T}L}{2\mathrm{π}2d}\right)$

This is a force on T due to b, and it is attractive.

$F_{Tc}=\left(\frac{{\mathrm{μ}}_0{i}_c{i}_{T}L}{2\mathrm{π}d}\right)$

This is a force on T due to C, and it is repulsive.

Hence, the total force acting on T is

$$\begin{gathered} F=F_{Tc}-F_{Tb} \\ F=\frac{2{\mathrm{μ}}_0{i}_{T}L}{4\mathrm{π}d}\left(i_c-\frac{i_b}{2}\right) \end{gathered}$$

Substitute all the values in the above equation.

$\begin{array}{rcl}F& =& \frac{2×\left(4\mathrm{π}×{10}^{-7} \text{T.m}/\text{A}\right)×5.0 \text{A}×3.0 \text{m}}{4\mathrm{π}×10×{10}^{-2} \text{m}}\left(3.2 \text{A}-\frac{5.0 \text{A}}{2}\right)\\ & =& 2.1×{10}^{-5} \text{N}\\ & & \end{array}$

The top wire is pushed by the central wire in the upward direction and pulled by the bottom wire in the downward direction.

Hence the force negative x-direction is $2.1×{10}^{-5} \text{N}$.