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Electromagnetic induction in neighbouring facilities

Short answer

AC current in a power line creates a magnetic field that links a parallel metallic facility, inducing voltage by mutual inductance. Example: 5 kA earth-fault current, 2 km parallel pipeline at 50 m, soil 100 Ωm induces about 1.84 kV, posing shock and corrosion risks.

Explain with example how electromagnetic induction occurs in neighboring facilities.

Electromagnetic (inductive) coupling is the voltage induced in a nearby metallic facility, such as a pipeline, fence, rail or telephone line, by the alternating magnetic field of the current in a power line. The facility and the power line are coupled through their mutual inductance, like the primary and secondary of an air-core transformer.

How it occurs

  1. The AC current II in the phase conductors sets up an alternating magnetic flux around the line.
  2. Part of this flux links the loop formed by the parallel facility and its earth return.
  3. By Faraday's law, a longitudinal emf is induced along the facility:
E=ωMIL  voltsE = \omega M I L\ \ \text{volts}

where MM is the mutual inductance per km, LL the length of parallel exposure (km) and II the inducing current. 4. With balanced load currents, the fields of the three phases largely cancel. The induction is therefore small in normal operation (it arises only from the unequal distances to each phase). 5. During a single line-to-ground fault, a large zero-sequence (earth-return) current flows and does not cancel. The induced voltage can then reach several kV.

Mutual inductance with earth return (Carson)

M=0.2ln⁡DeD mH/km,De=658.5ρf mM = 0.2\ln\frac{D_e}{D}\ \text{mH/km}, \qquad D_e = 658.5\sqrt{\frac{\rho}{f}}\ \text{m}

Example

A metal pipeline runs parallel to a 132 kV line for L=2L = 2 km at D=50D = 50 m. Take soil resistivity ρ=100 Ω\rho = 100\ \Omegam, f=50f = 50 Hz, and an earth-fault current of 5 kA.

De=658.5100/50=931.3 mM=0.2ln⁡931.350=0.585 mH/kmE=2π×50×0.585×10−3×5000×2=1838 V≈1.84 kV\begin{aligned} D_e &= 658.5\sqrt{100/50} = 931.3\ \text{m} \\ M &= 0.2\ln\frac{931.3}{50} = 0.585\ \text{mH/km} \\ E &= 2\pi \times 50 \times 0.585 \times 10^{-3} \times 5000 \times 2 \\ &= 1838\ \text{V} \approx 1.84\ \text{kV} \end{aligned}

This voltage would appear between the pipeline and remote earth, which is dangerous to anyone touching it.

Effects on neighbouring facilities

  • Shock hazard to people touching pipelines, fences or rails
  • Noise and interference in telephone and communication circuits
  • AC corrosion of pipelines and damage to their coatings and cathodic protection
  • Wrong operation of railway signalling

Remedies

  • Keep adequate separation and short parallel runs; cross at right angles.
  • Transpose the power line; use earth wires of low resistance (they carry part of the return current and reduce the net field).
  • Earth the facility at intervals, and use insulating joints and gradient-control mats.
  • Clear faults quickly with fast protection.

✓ Answer checked by Claude Opus 5.5 (high effort)

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