# Does magnetic field difference move current-carrying wire?

My high school book explains the magnetic force on current-carrying wire by stating:

The change in the resultant of the magnetic flux density at the two sides of the wire which is due to both the original magnetic flux and the magnetic flux due to the flow of the electric current in the wire

This force leads to the motion of the wire from the higher magnetic flux density region to the lower magnetic density region if the wire is free to move

It simply states that the wire move due to magnetic flux density difference. How could that even affect the wire!

Is this explanation wrong? If not, then what is the correct explanation?

• Can you specify the set up? – Hugo V Oct 18 '18 at 22:07

The force on a straight current ($$I$$) carrying wire (of length $$l$$) due to an external magnetic field $$\vec{B}$$ (vector) is $$\vec{F}=I\vec{l}\times\vec{B}$$ where $$\vec{l}$$ is the vector which has length $$l$$ and points in the direction of the current. This is basically the Lorentz force restricted to a situation of a straight current carrying wire. The direction of the force can be obtained with the right hand rule for cross products and is dependent on the direction of current flow and the direction of the magnetic field. Even if you have a uniform magnetic field (no areas of higher or lower "magnetic flux densities"), there would still be a Lorentz force present if that magnetic field is not aligned with the current. At face value, it appears that your textbook is just plain wrong, but I might be missing some context (e.g. a different set up than the one I'm imagining).