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From the following diagram:

enter image description here

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

enter image description here

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 FL than the one that has the magnetic field incompletely covering it.

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  • \$\begingroup\$ Is it a loop or single open wire ? \$\endgroup\$ – Plutonium smuggler May 31 '15 at 11:56
  • \$\begingroup\$ @Plutoniumsmuggler sorry for not replying earlier, both diagrams represent a single wire in a circuit(neglecting the PS and the connection to it). \$\endgroup\$ – Pupil Jun 7 '15 at 20:55
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Let start by defining the variables in your equation: \$ F=IL\times B \$

-I is the current, within your field

-L is the Length of your conductor in the direction of current flow

-B is the magnetic field (magnetic flux density, more correctly)

Due to the cross product and the right hand rule, we know that the direction of force is to the right.

You're intuition is correct, the force in the second diagram is less, but not because L is smaller, but rather because not all your current is within the influence of the field.

If we assume that the current is uniformly distributed throughout the width of the wire then the force on the wire will be proportional to the the width of the wire within the influence of the field.

That is, if 20% of the wire's width is within the field, then the force on the wire will be 20% as much as it would have been had the whole wire been in the field.

\$ F=\frac {w'}{w}F_0 \$

where \$F_0\$ is the force in the first diagram and \$ \frac {w'}{w} \$ is the ratio of the width of the wire within the field.

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  • \$\begingroup\$ Great, I assumed it would be less. But there is an interesting effect someone pointed out to me. How the wire would not be subjugated to motion! Or any net-force(on the wire, because the electrons are pushed from the region inside the magnetic field to the region without the magnetic field. \$\endgroup\$ – Pupil May 31 '15 at 18:14
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    \$\begingroup\$ @Key Hmmm - are you certain? Due to an action reaction pair I'd be very surprised if there was no net force on the conductor. The electrons can't escape the conductor after all. This is very similar to a Hall probe (Hall Effect). As the electrons experience a force perpendicular to the length of the wire a voltage/potential difference is created between the 2 sides of the wire! This potential difference stabilizes when the force of the Electric Potential is equal to the Lorentz force on the electron. \$\endgroup\$ – Rohan Jun 1 '15 at 8:00
  • \$\begingroup\$ OPS. I just read that post all wrong, never mind what I said earlier. Misunderstood... \$\endgroup\$ – Pupil Jun 1 '15 at 11:47

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