Magnetic Forces Between Parallel Current-Carrying Conductors

Expert reviewed • 22 November 2024 • 5 minute read

Introduction

When electric current flows through a conductor, it generates a magnetic field around it. When two current-carrying conductors are placed parallel to each other, their magnetic fields interact, resulting in either attractive or repulsive forces between them. This phenomenon forms the foundation for many electromagnetic devices and was historically used to define the ampere unit.

Magnetic Field Generation

A current-carrying wire produces a magnetic field that forms concentric circles around the wire. The direction of this magnetic field can be determined using the right-hand grip rule:

  • Point your right thumb in the direction of conventional current
  • Your curled fingers indicate the direction of the magnetic field lines

Force Between Parallel Conductors

When two parallel conductors carry current:

  • If currents flow in the same direction, the conductors attract each other
  • If currents flow in opposite directions, the conductors repel each other

The magnitude of this force is given by:

F=μ02πI1I2LrF = \frac{\mu_0}{2\pi} \frac{I_1I_2L}{r}

Where:

  • FF is the force (in Newtons)
  • μ0\mu_0 is the permeability of free space 4π×10−74\pi \times 10^{-7} T⋅m/A)
  • I1I_1 and I2I_2 are the currents in each wire (in Amperes)
  • LL is the length of the parallel section (in meters)
  • rr is the separation distance between the wires (in meters)

The SI Definition of Ampere

Historically, this interaction was used to define the ampere. The original definition stated:

One ampere is the constant current that, when maintained in two straight parallel conductors of infinite length, of negligible cross-section, and placed one meter apart in vacuum, produces a force between these conductors of 2×10−72 \times 10^{-7} newtons per meter of length.

Note: As of 2019, the ampere is now defined in terms of the elementary charge e=1.602176634×10−19coloumbse = 1.602176634 \times 10^{-19}\text{coloumbs}

Practice Question 1

Two parallel conductors, each 1.00 m long and separated by 0.500 m, carry currents of 12.0A in opposite directions. Calculate the repulsive force between them. Dtermine the produced force.

Solution: F=(4π×10−7)(12.0)(12.0)(1.00)2π(0.500)=5.76×10−5NF = \frac{(4\pi \times 10^{-7})(12.0)(12.0)(1.00)}{2\pi(0.500)} \\= 5.76 \times 10^{-5} N

The force is relatively small but measurable with sensitive equipment.

Applications

This principle has practical applications in:

  • Electromagnetic devices
  • Current-measuring instruments
  • Electric motors
  • Power transmission systems

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The Motor Effect: Forces on Current-Carrying Conductors

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