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From the following diagram:

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Lorentz force can be calculated from the following formula: $F$ = $IL\times B$ What about this wire in the following diagram?

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Assuming that all the variables are the same($I,L,B$). However, the diameter(range) of the magnetic field is much smaller. Would both conductors/wires experience the same Lorentz force? My assumption is the second diagram would have a much smaller $F_L$ than the one that has the magnetic field incompletely covering it.

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A Lorentz force can just be generated at places where the current and the magnetic field is non-zero. So the parts of the wire which are not immersed in the magnetic field don't give any contribution.

You can calculate the force by just going over to the continuous formulation of the Lorentz force law: $$ \vec{f}=\vec{j} \times \vec{B}$$

If all quantities are assumed to be homogeneous over the regions of space you indicated (especially the current over the wire) the force per length is given by: $\frac{\vec{F}}{L}=A \cdot \vec{j} \times \vec{B}$, where A is the area of a cut through the wire where $j\neq0$ and $B\neq 0$. (In your drawing this is the area which you can imagine as a horizontal line inside both boxes.)

So in the second case the force would be indeed smaller.

But there is another pretty essential effect taking place here: The Lorentz force is a force on electrons moving in the wire. So effectively the Lorentz force will push all moving electrons into the region where there is no magnetic field and the wire will not be subject to a force. (Assuming the right direction for the $B$-field. Check it yourself! For other directions of the field it might also cause that the wire is sucked in and will be completely covered by the magnetic field.)

edit: This will in turn lead to a increased charge density in the right part of the wire, so again: There is a force on the wire, effectively as large as previously calculated.

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  • $\begingroup$ What about induced $\epsilon$? Is it possible to have any induced voltage & current flow from the second diagram? $\endgroup$
    – Pupil
    May 31, 2015 at 18:10
  • $\begingroup$ I believe it is possible, because $\epsilon$ = -$vBL$ all the variables are non-zero. However, I believe the second diagram would induce the same voltage(assuming all variables $v,B,L$ are equal) however, the magnitude of induced current would also be smaller. But I'm not sure considering the effect you pointed out. $\endgroup$
    – Pupil
    May 31, 2015 at 18:10
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    $\begingroup$ Yes, I also realized that this is true: In the second picture you would induce an electric field due to accumulation of charge density in a certain region in the wire. And again there should be force onto the wire pushing it out. $\endgroup$ May 31, 2015 at 18:28
  • $\begingroup$ Thank you for that edit, I understood something incompletely different.However, is my point correct about induced EMF? That both conductors would induce the same voltage, however, the current in the first diagram is great. $\endgroup$
    – Pupil
    Jun 1, 2015 at 12:15
  • $\begingroup$ What effect are you thinking of? Hall effect? $\endgroup$ Jun 1, 2015 at 17:22

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