Answer :
72.828×[tex]10^{-7}[/tex]Volt is the Hall potential between the edges of the ribbon.
We know that Hall potential is given by =IB/qnd where
I is the current flowing
,B is the magnetic field ,
q is the charge of the electron,
n is the charge density of electron,
and d is the thickness
So, we are given that current(I)=110Ampere,magnetic field(B)=1.25Tesla,
charge of electron(q)=1.6×[tex]10^{-19}[/tex]C, charge density (n)= 5.9x1028,thickness
(t) as 2.cm
So, putting the values of the respected quantities on the above formula ,we get
Hall potential([tex]V_h[/tex])=(110×1.25)/(1.6×[tex]10^{-19}[/tex]×5.9×[tex]10^{28}[/tex]×0.2×[tex]10^{-2}[/tex])
=>([tex]V_h[/tex])=[137.5/1.888]×[tex]10^{-7}[/tex]
=>([tex]V_h[/tex])=72.828×[tex]10^{-7}[/tex]Volt.
Hence, the Hall potential is 72.828×[tex]10^{-7}[/tex]Volt.
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(Complete question) is:
Figure shows a silver ribbon whose cross section is 1.0 cm by 0.20 cm. The ribbon carries a current of 110 A from left to right, and it lies in a uniform magnetic field of magnitude 1.25 T. Using a charge density value of n = 5.9x1028 electrons per cubic meter for silver, find the Hall potential between the edges of the ribbon.
