Physics 12.13 - Force on a current-carrying conductor

Physics 12.13 - Force on a current-carrying conductor

Calculate the force on a perpendicular current-carrying conductor, convert centimetres and millinewtons, and rearrange the equation to find field strength, current or the length within the field. This is Higher-tier content.

Force, field, current and length

Fleming’s left-hand rule predicts force direction. To calculate the magnitude for a straight conductor at right angles to a uniform magnetic field, use:

Force on a perpendicular conductor

F=BIlF=B I l
SymbolQuantitySI unit
FFForce on the conductornewton, N\mathrm{N}
BBMagnetic flux density: a measure of field strengthtesla, T\mathrm{T}, equivalent to N A−1 m−1\mathrm{N}\,\mathrm{A}^{-1}\,\mathrm{m}^{-1}
IIConventional currentampere, A\mathrm{A}
llLength of conductor inside the fieldmetre, m\mathrm{m}

The symbol ll is a lower-case L, not the number 1. A wire may be long, but only its section within the magnetic field contributes to this calculation. The equation above applies when the wire and field are perpendicular; do not use it as the full force for any angle.

For fixed field and length, doubling current doubles force. Similarly, doubling either BB or ll alone doubles force. Reversing current changes force direction, but keeping its magnitude unchanged keeps the same force magnitude.

Converting units and calculating force

Worked example: finding force

A 6.0 cm6.0\,\mathrm{cm} wire section is perpendicular to a field of flux density 0.45 T0.45\,\mathrm{T}. It carries 2.8 A2.8\,\mathrm{A}. Calculate the force.

Convert the length before substituting:

l=6.0100=0.060 ml=\frac{6.0}{100}=0.060\,\mathrm{m} F=0.45×2.8×0.060=0.0756 NF=0.45\times2.8\times0.060=0.0756\,\mathrm{N}

To two significant figures, the force is 0.076 N0.076\,\mathrm{N}. Using 6.0 as the length in metres would give a force 100 times too large. A small force is reasonable for this short wire.

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