Potential difference between plates in a B-Field?

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SUMMARY

The discussion centers on calculating the potential difference between two plates in a magnetic field (B-field) when a charged particle moves between them. It is established that the magnetic field is non-conservative, which complicates the use of potential energy methods. The charge's negative sign is determined using the right-hand rule, indicating the direction of the electric field between the plates is upward, opposing the downward force exerted by the B-field. The potential difference (V) can be calculated using the formula V = Ed, where E is derived from the equation E = vB, leading to V = vBd.

PREREQUISITES
  • Understanding of electromagnetic theory, specifically the behavior of charges in magnetic fields.
  • Familiarity with the right-hand rule for determining charge motion in magnetic fields.
  • Knowledge of electric fields and their relationship to potential difference.
  • Basic proficiency in algebraic manipulation of equations related to electromagnetism.
NEXT STEPS
  • Study the derivation and applications of the Lorentz force equation, F = q(E + v × B).
  • Learn about the implications of non-conservative fields in electromagnetism.
  • Explore the concept of electromagnetic induction and its relation to potential difference.
  • Investigate the relationship between electric and magnetic fields in the context of Maxwell's equations.
USEFUL FOR

Students and professionals in physics, particularly those focusing on electromagnetism, electrical engineering, and anyone involved in the study of charged particles in magnetic fields.

thursdaytbs
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How is potential difference found between two plates in a B-field? Any help is greatly appreciated.
 
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What potential are u talking about?Magnetic potential...?

Daniel.
 
Unfortunately, the magnetic field cannot be treated by the method of potential energy. This is because the magnetic field is not a conservative field (it depends on the speed you are going and the path you take, unlike a gravitational or electrostatic field).
 
Well I'm given two plates and a B-field. a charge is placed inbetween the two plates and moves paralllel to the two plates until the end, where it then makes a downward circle turn. All the while, the charge is in a B-Field.

the distance between the two plates is given
the B-field is given
the radius of the downward circle is given
and the velocity of the charge is given.
 
The question runs me through a series of questions actually, and I think they are all linked somehow.

The first one asks what the sign of the charge of the particle is. And the answer I got was Negative, since the B-Field is going INTO the paper, and the velocity is to the right. Therefore, the charge SHOULD be moving upwards [right hand rule number one], but instead it moves downwards, therefore the charge is negative.

The second question is indicate the direction of the electric field between the plates. My repsonse was the electric field moves from the bottom plate to the top plate, because the charge is being pushed downward by the B-Field [seen when the charge exits the two plates]. Since it should be moving downwards, and is not [the charge moves in a straight parallel line between the two plates]. Therefore, the plates electric field must be opposing the downward force with an upward force. Aka a force from the bottom plate to the top plate.

And the third question asks what's the magnitude of the potential difference (V) between the plates, which is where I got stuck.

On second thought. emf = BLv, where B = b-field, L = distance between the plates, and v = the velocity of the charge. Would this solve for the potential difference between the plates? If so, how come? any simple reason?
 
Last edited:
Equate the electric force (in ansolute value) with the magnetic one and from there extract the value of E.Then V=Ed.

Daniel.
 
F = Eq = qvB. Therefore E = vB
So.. V=Ed = vBd?
 
Awsome, thanks so much.
 

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