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Chapter 4: Q50E (page 915)

Let Fig. E27.49 represent a strip of an unknown metal of the same dimensions as those of the silver ribbon in Exercise 27.49. When the magnetic field is 2.29 T and the current is 78.0 A, the Hall emf is found to be 131 µV. What does the simplified model of the Hall effect presented in Section 27.9 give for the density of free electrons in the unknown metal?

Short Answer

Expert verified

Free electrons per cubic metern=3.7×1028m−3

Step by step solution

01

Drift velocity

The drift velocity is given by

vd=Jxn|q|

02

Determine the drift velocity

The drift velocity is

vd=Jxn|q|

Jx=IA=Iy1z1vd=In|q|y1z1

Magnetic force is equal to the electric force

|q|Ez=|q|vdByvd=EzBy

Form these equations we find that

n=IByy1|q|E=(78.0A)(2.29T)(2.3×10−4m)(1.6×10−19C)(1.31×10−4V)=3.7×1028m−3

Therefore, simplified model of hall effect electrons per cubic meter is

n=3.7×1028m−3

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