When calculating the Lorentz force of a wire the formula used is:

$$ F = IL \times B$$

I'm curious about the distribution of current($I$) in correlation to the magnitude of the force ($F_L$).

If I applied direct current to a wire and placed it near a magnet, there is a force.

enter image description here

The diagram above is a top view of an example.

The magnet and it's field, and the wire(current going out the page). I learned that with the case of DC, I can assume the current is distributed evenly. However,I'm confused about the nature of how this works... Does that mean that $I$ is the same value at points a,b,c or any point arbitrary point in the conductor? I know that the magnetic field isn't since it weakens with the increase of $r$.

It's a bit counter intuitive for me, to imagine how $1A$ is the value of $I$ in all points.

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    $\begingroup$ The 1A is kind of a total across all the points that make up the cross-section of the wire. $\endgroup$ Aug 2, 2016 at 3:13

1 Answer 1


To expand on David Wallace's comment, the current $I$ is actually the sum of the current density across the cross-section of the wire. The current density is related to the electric field in the wire (which is related to the voltage of your power source). Mathematically,

$I=\int \vec J \cdot \vec {dA}$

$\vec J=\sigma \vec E$

Current density $\vec J$ is the current per unit area, $\sigma$ is the conductivity of the wire (inverse of the resistivity $\rho$). So the assumption that the current is distributed evenly across a wire in the DC case means that $\vec J$ is equal at every point in the wire, not that $1 A$ is flowing through every point in the wire.


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