Comparing Hall Voltage of Bars 1 & 2

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The discussion centers on comparing the Hall voltage of two identical metal bars, where bar 2 is twice as wide as bar 1, both carrying the same current in a uniform magnetic field. The Hall effect equation is referenced, indicating that the Hall voltage is related to the electric field and the magnetic field. It is concluded that the Hall voltage of bar 2 will be twice that of bar 1 due to its increased width, as the electric field (E) is directly related to the voltage (V) and inversely related to the distance (d). There is some confusion about the necessity of using the Hall effect equation when the relationship can be established through E=V/d. Understanding the Hall effect is essential for accurately analyzing the voltage differences in this scenario.
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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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