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Two long parallel conductors attract for currents in the same direction and repel for currents in opposite directions.
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Formula
F/ℓ in N/m; μ0 in N/A2; I1 and I2 in A; d in m.
| Symbol | Meaning | Unit |
|---|---|---|
| Force per conductor length | N/m | |
| Magnetic constant | N/A2 | |
| Currents in the parallel conductors | A | |
| Distance between conductor axes | m |
\frac{F}{\ell}=\frac{\mu_0 I_1I_2}{2\pi d}The relation combines the field of a long straight conductor with the force on a second current-carrying conductor. It applies to very long parallel wires in vacuum or approximately air, when separation is small relative to wire length and large relative to wire radii. Currents in the same direction attract; opposite currents repel. End effects and nearby ferromagnetic materials are excluded.
The relation combines the field of a long straight conductor with the force on a second current-carrying conductor. It applies to very long parallel wires in vacuum or approximately air, when separation is small relative to wire length and large relative to wire radii. Currents in the same direction attract; opposite currents repel. End effects and nearby ferromagnetic materials are excluded.
Two long parallel conductors attract for currents in the same direction and repel for currents in opposite directions.
Force per conductor length
Long parallel conductors; direction depends on current orientation.
Calculate the quantity from the worked example for “Force Between Parallel Currents”.
I1=5 A, I2=10 A and d=5.0 cm give F/ℓ=(2×10−7)(50)/0.050=2.0×10−4 N/m.
Over 2.0 m, F=4.0×10−4 N.
Check: doubling the separation halves force per length.
Ignoring units or assumptions stated for the formula: F/ℓ in N/m; μ0 in N/A2; I1 and I2 in A; d in m.
Convert units before substitution and verify the validity range; round only at the end.
Formula cluster
Related relations: magnetische-kraft-leiter, magnetfeld-langer-gerader-leiter.
I1=5 A, I2=10 A and d=5.0 cm give F/ℓ=(2×10−7)(50)/0.050=2.0×10−4 N/m. Over 2.0 m, F=4.0×10−4 N. Check: doubling the separation halves force per length.
Estimate magnetic interactions between current-carrying wires.
Here is how to work through a typical Force Between Parallel Currents (F/ℓ=μ₀I₁I₂/(2πd)) task step by step:
Task
Calculate the quantity from the worked example for “Force Between Parallel Currents”.
Solution path
I1=5 A, I2=10 A and d=5.0 cm give F/ℓ=(2×10−7)(50)/0.050=2.0×10−4 N/m.
Over 2.0 m, F=4.0×10−4 N.
Check: doubling the separation halves force per length.
The relation for Force Between Parallel Currents connects measurable quantities and shows which variables determine the result. Two long parallel conductors attract for currents in the same direction and repel for currents in opposite directions. First define the state and reference used. The equation is more than a shortcut: its proportionalities let you predict changes without numerical calculation. Ask whether doubling an input doubles the result, or whether an angle, area or squared distance changes that dependence. This qualitative check helps catch an incorrect setup early and gives a physical interpretation to the number obtained. It also clarifies which measurement needs the greatest care.
I1=5 A, I2=10 A and d=5.0 cm give F/ℓ=(2×10−7)(50)/0.050=2.0×10−4 N/m. Over 2.0 m, F=4.0×10−4 N. Check: doubling the separation halves force per length. A reliable calculation lists known quantities with units, rearranges for the unknown, and rounds only at the end. Compare the output unit and magnitude with the described process; the equation alone cannot tell whether the input data are sensible. If a result seems surprising, check signs, unit conversions and angle definitions first. A limiting-case check helps too: imagine setting a parameter to zero or doubling it, then ask whether the result responds as the model predicts. In an experiment, record measurements separately from assumptions so discrepancies can be traced to arithmetic or idealisation.
The relation combines the field of a long straight conductor with the force on a second current-carrying conductor. It applies to very long parallel wires in vacuum or approximately air, when separation is small relative to wire length and large relative to wire radii. Currents in the same direction attract; opposite currents repel. End effects and nearby ferromagnetic materials are excluded. If those assumptions fail, simply substituting the same numbers is not enough; a more general model or additional terms are needed. Pay particular attention to signs, reference frame and the definition of each quantity. Measurements can depart from the ideal because real bodies, fields or processes are not perfect. State the assumption that matters for your calculation; this helps distinguish a substitution error from a model choice. A limiting case that violates the assumptions is not a reason to extend the formula blindly.
Write the units beside every measured value and convert them before substitution. Then check that both sides of the relation have matching dimensions; for example, a velocity must have length per time, whereas energy must be in joules. I1=5 A, I2=10 A and d=5.0 cm give F/ℓ=(2×10−7)(50)/0.050=2.0×10−4 N/m. Over 2.0 m, F=4.0×10−4 N. Check: doubling the separation halves force per length. gives a concrete scale check: compare the result with the physical situation and ask whether its magnitude is plausible. Avoid rounding intermediate values too early, because small rounding errors can grow through powers, roots or very small quantities. Correct dimensions do not prove a calculation, but they rule out many wrong rearrangements and unit mistakes.
Estimate magnetic interactions between current-carrying wires. In a problem this relation often supplies only part of the description: another equation may add a missing measurement, a conservation balance or a boundary condition. First define the system and list the quantities known for it. Then use each formula only under its own assumptions; the same symbol can play different roles in different models. Write the combined reasoning in clear steps. This makes it possible to see which conclusion comes from Force Between Parallel Currents and what information the additional relation contributes.
Quanta exam set
Curated exam set for "Force Between Parallel Currents":
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Which formula describes Force Between Parallel Currents?
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How do you rearrange Force Between Parallel Currents for Force per conductor length?
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Which common mistake happens with Force Between Parallel Currents?
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