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Fields Around Parallel Conductors

Exam code: 5054
Written by: Ashika|Reviewed by: Caroline Carroll|Updated 2 July 2026

Fields Around Parallel Wires

Fields Around Parallel Wires

  • A current carrying conductor, such as a wire, produces a magnetic field around it

  • The direction of the field depends on the direction of the current through the wire

    • This is determined by the right hand thumb rule

  • Parallel current-carrying conductors will therefore either attract or repel each other

    • If the currents are in the same direction in both conductors, the magnetic field lines between the conductors cancel out – the conductors will attract each other

    • If the currents are in the opposite direction in both conductors, the magnetic field lines between the conductors push each other apart – the conductors will repel each other

Repulsion & Attraction of Current-Carrying Wires

20.1 Same or opposite direction current_2

Both wires will attract if their currents are in the same direction and repel if in opposite directions

  • When the conductors attract, the direction of the magnetic forces will be towards each other

  • When the conductors repel, the direction of the magnetic forces will be away from each other

  • The magnitude of each force depends on the amount of current and the length of the wire

 

Force per Unit Length Between Two Parallel Conductors

  • FL{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true} is the force per unit length between two parallel currents I1 and I2 separated by a distance r 

  • The force is attractive if the currents are in the same direction and repulsive if they are in opposite directions

  • It is calculated using the equation:

FL = μ0 I1I22πr{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true}

  • Where:

    • F is the force applied between the two parallel wires (N)

    • L is the length of each parallel conductor (m)

    • μ0 is the constant for the magnetic permeability of free space = 4π × 10−7 N A−2

    • I1 is the current through the first conducting wire (A)

    • I2 is the current through the second conducting wire (A)

    • r is the separation between the two conducting wires (m)

Forces on Current-Carrying Wires

T_fIGs2p_4-2-force-eqn-explanation

The forces on each of the current-carrying wires are equal and opposite in direction

Obtaining the Equation

  • The force from wire 2 on wire 1, F2 = B2I1Lsin(θ) 

  • In this situation the magnetic field is perpendicular to the current in the wire, so sin(θ) = 1

  • F2 = −F1 so the force between them is F

  • The force on a unit length of the wires is then given by FL = B2I1LL{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true}

    • Hence, FL = B2I1{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true}

  • The magnitude of the magnetic field at a radial distance, r away from the current conducting wire is: B = μ0I2πr{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true}

    • In this case the magnetic field strength from B2 at a distance r away from wire 2 is: B2 = μ0I22πr{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true}

  • Substituting for B2 into the force per unit length equation gives us: FL = μ0I22πrI1{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true}