Question

What is the net force (magnitude and direction) on each wire in FIGURE EX29.36?

Short answer

The net force (magnitude and direction) on each wire is $$\begin{gathered} B_1=5\times {10}^{-5}T \\ \overrightarrow{F_1}=25\times {10}^{-5}N \end{gathered}$$

Step-by-step solution

Given Information

We need to find the net force on each wire given in the diagram.

Simplify

Consider the given values:

$I_1=10A$

$I_2=10A$

$I_3=10A$

$l=0.5m$

The net force (magnitude and direction) on each wire on the diagram is

Calculation

For l$\Rightarrow Bat1$

$B_1=\frac{{\mu }_0}{2\Omega }\times \frac{l_2}{\left(2\times {10}^{-2}\right)}-\frac{{\mu }_0}{2\Omega }\times \frac{l_2}{\left(4\times {10}^{-3}\right)}$

$B_1=\left(2\times {10}^{-7}\right)\left(\frac{10}{2\times {10}^{-2}}-\frac{10}{4\times {10}^{-2}}\right)$

$B_1=5\times {10}^{-5}T$

role="math" localid="1649527850039">F1→=I1lB1

$\overrightarrow{F_1}=10\times 0.5\times 5\times {10}^{-5}$

$\overrightarrow{F_1}=25\times {10}^{9-5}$