Conductor moving through a mag. field at constant speed doesn’t build up charge?

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Maya$tudent
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I’m doing a past paper, Edexcel IAL January 2014 unit 4, question 13.
Part b(iii)

I don’t understand why the electric force produces by the build up of charge acting opposite to the magnetic force would cause no further charge build up if the speed is constant...

Please help 🙏🙏
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Eimmanuel
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(Original post by Maya$tudent)
I’m doing a past paper, Edexcel IAL January 2014 unit 4, question 13.
Part b(iii)

I don’t understand why the electric force produces by the build up of charge acting opposite to the magnetic force would cause no further charge build up if the speed is constant...

Please help 🙏🙏
You need to know there are a few things that are happening as a conductor is moving perpendicular to the magnetic field. I would use the diagram below to explain what is going on.
Name:  Motion_emf_01.jpg
Views: 30
Size:  52.9 KB


Initially, there is no separation of charges in the conductor. As the moving conductor is “cutting” the magnetic field, the mobile electrons in the conductor are considered to be moving (to the right) with respect to the magnetic field. As a result, the mobile electrons experience a magnetic force (FM = qvB) and moves towards the upper-end A, creating an excess of negative charge at the upper-end A and excess of positive charge at the lower end B.

Because of the separation of charges, a net electric field is produced inside the conductor. This electric field would cause the mobile electrons that are NOT at the upper-end A to experience an electric force (FE = qE) downward. The magnitude of the electric field depends on the amount of charges at the opposite ends.

Note that the magnetic force is still exerting on these mobile electrons and the magnitude of the magnetic force is constant. The magnetic force is greater than that of the electric force, so the charge continues to accumulate at the ends of the conductor. There will be a time when the electric field becomes large enough for the downward electric force (with magnitude qE) to cancel exactly the upward magnetic force (with magnitude qvB). Then qE = qvB and the charges are in equilibrium.

Hope it makes sense to you.
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Maya$tudent
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(Original post by Eimmanuel)
You need to know there are a few things that are happening as a conductor is moving perpendicular to the magnetic field. I would use the diagram below to explain what is going on.
Name:  Motion_emf_01.jpg
Views: 30
Size:  52.9 KB


Initially, there is no separation of charges in the conductor. As the moving conductor is “cutting” the magnetic field, the mobile electrons in the conductor are considered to be moving (to the right) with respect to the magnetic field. As a result, the mobile electrons experience a magnetic force (FM = qvB) and moves towards the upper-end A, creating an excess of negative charge at the upper-end A and excess of positive charge at the lower end B.

Because of the separation of charges, a net electric field is produced inside the conductor. This electric field would cause the mobile electrons that are NOT at the upper-end A to experience an electric force (FE = qE) downward. The magnitude of the electric field depends on the amount of charges at the opposite ends.

Note that the magnetic force is still exerting on these mobile electrons and the magnitude of the magnetic force is constant. The magnetic force is greater than that of the electric force, so the charge continues to accumulate at the ends of the conductor. There will be a time when the electric field becomes large enough for the downward electric force (with magnitude qE) to cancel exactly the upward magnetic force (with magnitude qvB). Then qE = qvB and the charges are in equilibrium.

Hope it makes sense to you.
This is so extremely helpful!!!! Thank you so much!
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Eimmanuel
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(Original post by Maya$tudent)
This is so extremely helpful!!!! Thank you so much!
Good luck to your exam.
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