Comparing Hall Voltage of Bars 1 & 2

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SUMMARY

The discussion centers on the comparison of Hall voltage between two identical metal bars, where bar 2 has twice the width of bar 1. According to the Hall effect equation, the Hall voltage is influenced by the width of the conductor. The conclusion drawn is that the Hall voltage of bar 2 will be twice that of bar 1 due to the relationship defined by E=V/d, where the electric field E is directly proportional to the voltage V and inversely proportional to the distance d. The use of the Hall effect is necessary to understand the underlying principles governing the voltage differences in magnetic fields.

PREREQUISITES
  • Understanding of the Hall effect and its equation: nq=(-JB/E)
  • Knowledge of electric fields and voltage relationships: E=(V/d)
  • Familiarity with the concepts of current and magnetic fields
  • Basic principles of electromagnetism
NEXT STEPS
  • Study the derivation of the Hall effect equation in detail
  • Explore the implications of varying conductor dimensions on Hall voltage
  • Investigate applications of the Hall effect in sensors and devices
  • Learn about the effects of magnetic field strength on Hall voltage
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in the practical applications of the Hall effect in electronic devices.

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Homework Statement



Two metal bars, 1 and 2, are identical in all ways, except that bar 2 has twice the width of 1. The bars are parallel to each other, but far apart from each other, in a uniform magnetic field and carry the same amount of current in a direction perpendicular to the field. How does the Hall voltage of bar 2 compare to that of bar 1?

Homework Equations


The equation for Hall effect is nq=(-JB/E). I know that I can relate E to V by using E=(V/d).


The Attempt at a Solution


If the width of bar 2 is twice the with of bar 1, then (because of E=V/d) the voltage of 2 would be two times greater than bar 1. I'm confused about why I need to use the Hall effect if I can just use E=V/d.
 
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