Three long wires all lie in an xyplane parallel to the xaxis. They are spaced equally,10 cm apart. The two outer wires each carry a current of 5.0 Ain the positive xdirection. What is the magnitude of the force on a3.0 m section of either of the outer wires if the current in the center wire is 3.2 A(a) in the positive xdirection and (b) in the negative xdirection?

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,$d=10\times {10}^{-2} \text{m}$.

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,

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

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

Similarly, we can write,

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

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

Hence, the total force acting on T is,

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

Substitute all the values in the above equation.

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

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" $1.7\times {10}^{-4} \text{N}$.

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

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

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

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

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

Hence, the total force acting on T is

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

Substitute all the values in the above equation.

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

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 $2.1\times {10}^{-5} \text{N}$.