A strip of copper placed in a magnetic field

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SUMMARY

The discussion focuses on calculating the Hall potential difference (V) across a copper strip measuring 10 mm in width and 700 micrometers in thickness, placed in a uniform magnetic field of 0.9 T. A current of 45 A flows through the strip, generating a Hall voltage due to the separation of charge carriers caused by the magnetic field. The key relationship established is that the Hall voltage arises when the magnetic force on the charge carriers equals the electric force generated by the separation of charges. The participants clarify that the cross-sectional area is determined by the width and thickness of the strip, which is essential for calculating the electric field (E).

PREREQUISITES
  • Understanding of Hall Effect and Hall voltage
  • Knowledge of magnetic fields and forces on charge carriers
  • Familiarity with electric fields and potential difference calculations
  • Basic principles of current flow in conductive materials
NEXT STEPS
  • Study the derivation of Hall voltage formula in conductive materials
  • Learn about the relationship between magnetic fields and electric currents
  • Explore the concept of charge carrier density in metals
  • Investigate applications of Hall Effect in sensors and devices
USEFUL FOR

Students in physics or electrical engineering, educators teaching electromagnetism, and professionals working with magnetic sensors and materials. This discussion is particularly beneficial for those looking to deepen their understanding of the Hall Effect and its applications.

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Homework Statement



A strip of copper 10 mm wide and 700 micrometers thick is placed in a uniform mag. field B magnitude 0.9 T, with B perpendicular to the strip. A current i = 45 A is then sent through the strip, such that a Hall Potential difference V appears across the width of the strip. Calculate V.

Homework Equations





The Attempt at a Solution



I know that V = Ed, so I think my first step is to find E.

However, I'm running into problems because it seems like they have given me a two dimensional object...

For example, I assume E = i/pA would work here to find E, but what exactly is cross sectional area if the object is a "two dimensional" strip?

I don't understand how they want me to work this. Any ideas?
 
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The strip has the given width and the given thickness (i.e. the cross-sectional area) and some length which is not given.
 
I see, but then how do I use V=Ed?

I assume this d is length?

Or should I rethink the approach entirely?

thanks.
 
Are you familiar with how a Hall voltage is generated?
 
Yes, the magnetic field is perpendicular to the strip which gives a force perpendicular to itself and the velocity of the charge carriers which tends to pull the charge carriers to one side of the strip.
 
As the charge cariers are passing along the length of the strip and through the cross-section of the strip, they are pulled one way by the magnetic field and so they pile up on one face of the strip. This separation of charges creates an electric field which will force the charge cariers in the opposite direction.
So the Hall voltage stops growing when the
magnetic force = electric force
Why not start from this step?
 

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