8. Force between two parallel current carrying conductors
Force between Two Parallel Current-Carrying Wires
Consider two parallel wires, \( a \) and \( b \). A current \( I_a \) passes through wire \( a \) and a current \( I_b \) passes through wire \( b \). The distance between them is \( d \).
The current \( I_a \) creates a magnetic field \( B_a \) that is perpendicular to \( I_b \) and its direction is away from \( I_a \).
Due to \( B_a \), a current element of wire \( b \) experiences a force \( dF_{ba} \) where:
\[ \vec {dF_{ba}} = I_b \, \vec {dl} \times \, \vec {B_a} \]
where \( dF_{ba} \) is the force on the current element of \( I_b \) by the wire carrying current \( I_a \). The direction of \( dF_{ba} \) will be towards wire \( a \).
The force on wire \( b \) per unit length is:
\[ \frac{dF_{ba}}{dl} \]
If wire \( b \) has a length of \( L \), the total force on wire \( b \) by \( I_a \) is:
\[ F_{ba} = \frac{\mu_0 I_a I_b L}{2 \pi d} \]
Similarly, wire \( a \) experiences an attractive force by wire \( b \) due to the current \( I_b \) flowing through it. \( I_b \) creates a magnetic field \( B_b \) that is towards the observer and passes through wire \( a \).
The force on the current element of wire \( a \) is:
\[ \vec dF_{ab} = I_a \, \vec {dl} \times \, \vec{B_b} \]
The force per unit length is the same, but their directions are opposite. Thus, the forces between two current-carrying wires follow Newton’s third law of motion.
Parallel wires carrying currents in the same direction attract each other, while parallel wires carrying currents in opposite directions repel each other. However, the magnitude of these forces remains the same and is given by:
\[ \frac{F}{L} = \frac{\mu_0 I_a I_b}{2 \pi d} \]
Definition of Ampere
Ampere is the value of the steady current which, when maintained in each of the two very long straight parallel wires of negligible cross-sections and placed one meter apart in vacuum, would produce on each of these wires a force of \( 2 \times 10^{-7} \) N per meter of length.
