Question

Derive the formula for the force acting between two parallel current carrying conductors.

Answer 1

Two current carrying straight conductors placed near each other will exert (magnetic) forces on each other due to magnetic field of each other.

Figure shows two long parallel conductors separated by a distance d and bearing currents $i_a$ and $i_b$. The conductor ‘$a$’ produces a magnetic field $B_a$ at all points along the conductor ‘$b$’. The right-hand rule tells us that the direction of this field is downwards. Its magnitude is given by,

$B_a=\frac{{\mu }_o{i}_a}{2\pi d}$

The conductor ‘$b$’ will experience a sideways force on account of the external field $B_a$. The direction of this force is towards the conductor ‘$a$’. you can verify this either by the cross product rule of vectors or by Fleming’s left hand rule. We label this force $F_{ba}$, the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by,

$F_{ba}=i_{b}L{B}_a$

$\frac{F_{ba}}{L}=\frac{{\mu }_o{i}_a{i}_b}{2\pi d}$

From considerations similar to above we can find that the force $F_{ab}$, on a segment of length $L$ of ‘$a$’ due to the current in ‘$b$’ is equal in magnitude to ‘$b$’ and directed towards ‘$b$’.