Question
Three coplanar parallel wires, each carrying a current of 10A along the same direction, are placed with a separation 5.0cm between the consecutive ones. Find the matnitude ol the magnetic force per unit lenght acting on the wires.
✓
Answer
i = 10A 
Magnetic force due to two parallel Current Carrying wires.$\text{F}=\frac{\mu_0\text{I}_1\text{I}_2}{2\pi\text{r}}$
So, $\overrightarrow{\text{F}}\ \text{or}\ 1=\overrightarrow{\text{F}}\ \text{or}\ 2+\overrightarrow{\text{F}}\ \text{by}\ 3$$=\frac{\mu_0\times10\times10}{2\pi\times5\times10^{-2}}+\frac{\mu_0\times10\times10}{2\pi\times10\times10^{-2}}$
$=\frac{4\pi\times10^{-7}\times10\times10}{2\pi\times5\times10^{-2}}+=\frac{4\pi\times10^{-7}\times10\times10}{2\pi\times10\times10^{-2}}$
$=\frac{2\times10^{-3}}{5}+\frac{10^{-3}}{5}=\frac{3\times10^{-3}}{5}=6\times10^{-4}\text{N}$ towards middle wire
Magnetic force due to two parallel Current Carrying wires.$\text{F}=\frac{\mu_0\text{I}_1\text{I}_2}{2\pi\text{r}}$So, $\overrightarrow{\text{F}}\ \text{or}\ 1=\overrightarrow{\text{F}}\ \text{or}\ 2+\overrightarrow{\text{F}}\ \text{by}\ 3$$=\frac{\mu_0\times10\times10}{2\pi\times5\times10^{-2}}+\frac{\mu_0\times10\times10}{2\pi\times10\times10^{-2}}$
$=\frac{4\pi\times10^{-7}\times10\times10}{2\pi\times5\times10^{-2}}+=\frac{4\pi\times10^{-7}\times10\times10}{2\pi\times10\times10^{-2}}$
$=\frac{2\times10^{-3}}{5}+\frac{10^{-3}}{5}=\frac{3\times10^{-3}}{5}=6\times10^{-4}\text{N}$ towards middle wire
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