Obtain equation for force between two parallel currents carrying wire.
Obtain equation for force between two parallel currents carrying wire.
Two long wire $a$ and $b$ are placed at $d$ distance apart from each other and $\mathrm{I}_{a}$ and $\mathrm{I}_{b}$ current
The conductor ' $a$ ' produces the same magnetic field $\overrightarrow{\mathrm{B}}_{a}$ at all points along the conductor ' $b$ '. The right hand rule tells us that the direction of this field is downwards. (When the conductor are placed horizontally).
$\mathrm{B}_{a}=\frac{\mu_{0} \mathrm{I}_{a}}{2 \pi d}$
$\ldots$ $(1)$
The conductor ' $b$ ' carrying a current $\mathrm{I}_{b}$ will experience a sideways force due to the field $\mathrm{B}$. The direction of this force is towards the conductor ' $a$ '.
Force on wire is given by,
$\overrightarrow{\mathrm{F}}=\mathrm{I} \vec{l} \times \overrightarrow{\mathrm{B}}$
$\mathrm{F}_{b a}=\left(\mathrm{I}_{b} \mathrm{~L}\right) \mathrm{B}_{a}$
$\mathrm{~F}_{b a}=\frac{\mu_{0} \mathrm{I}_{a} \mathrm{I}_{b} \mathrm{~L}}{2 \pi d}$ $...(2)$
It is of course possible to compute the force on ' $a$ ' due to ' $b$ '. From consideration similar to above we can find the force $\mathrm{F}_{a b}$ on a segment of length $\mathrm{L}$ of ' $a$ ' due to the current in ' $b$ '. It is equal in magnitude to $\mathrm{F}_{b a}$ and directed towards ' $\mathrm{b}$ '. Thus,
$\overrightarrow{\mathrm{F}}_{b a}=-\overrightarrow{\mathrm{F}}_{a b}$
$\ldots$ $(3)$
This is consistent with Newton's third law.
We have seen from above that currents flowing in the same direction attract each other. One can show that oppositely directed currents repel each other.
Thus, parallel currents attract and antiparallel currents repel.
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