A very thin rectangular strip of zinc has been deposited on an insulating substrate. Let the length, width, and thickness of the strip be x, y, and z, respectively. The length and width are measured to be 10.0 cm and 2.0 cm. When a potential difference of 20 mV is applied as shown in Figure 11.9a

Question

A very thin rectangular strip of zinc has been deposited on an insulating substrate. Let the length, width, and thickness of the strip be x, y, and z, respectively. The length and width are measured to be 10.0 cm and 2.0 cm. When a potential difference of 20 mV is applied as shown in Figure 11.9a (page 404), the current through the strip is 400 mA.

(a) Use the fact that the resistivity of zinc is (at room temperature) [latex] ho=5.92 imes 10^{-8} \Omega \cdot m[/latex] to find the thickness of the strip.

(b) Now a magnetic field of 0.25 T is applied perpendicular to the strip as shown in Figure 11.9b. A Hall voltage [latex]V_{ H }=+0.56 \mu V[/latex] appears when the voltmeter leads (+ and -) are connected as shown. Determine the sign of the charge carriers and their number density.

Strategy (a) The resistance of a wire with uniform crosssectional area is related to the resistivity by [latex]R= ho x / A[/latex], where A is the cross-sectional area and x is the length of the wire. In this case [latex]A=y z[/latex]. This allows us to find the thickness of a wire with known resistivity.

(b) In equilibrium the magnetic force on a charge carrier is equal to the electric force due to the Hall voltage. Thus [latex]e E=e v B[/latex]. From basic electrical conductivity, the electron drift speed is [latex]v=I / n e A[/latex]. These two relations allow us to solve for the charge carrier density n.

Accepted Answer

(a) The resistance of the sample is [latex]R= ho x / A[/latex]. From Ohm’s law, [latex]R=V / I, ext { so } R=V / I= ho x / y z[/latex]. Therefore

[latex]z=\frac{ ho x I}{y V}=\frac{\left(5.92 imes 10^{-8} \Omega \cdot m ight)(0.10 m )(0.400 A )}{(0.02 m )(0.02 V )}[/latex]

[latex]=5.92 imes 10^{-6} m[/latex] (11.5)

(b) Combining the relations [latex]e E=e v B ext { and } v=I / n e A[/latex],

[latex]\frac{e V_{ H }}{y}=e B \frac{I}{n e A}[/latex] (11.6)

where we have used the fact that for this geometry [latex]E=V_{ H } / y[/latex]. Finally, because [latex]A=y z[/latex], Equation (11.6) reduces to

[latex]n=\frac{I B}{e V_{ H } z}[/latex] (11.7)

[latex]\begin{aligned}&=\frac{(0.400 A )(0.25 T )}{\left(1.6 imes 10^{-19} C ight)\left(5.6 imes 10^{-7} V ight)\left(5.92 imes 10^{-6} m ight)} \&=1.89 imes 10^{29} m ^{-3}\end{aligned}[/latex]

This is a very high density (slightly higher, in fact, than one finds for copper). It turns out that the higher the density of conductors (electrons or holes), the thinner the strip needs to be to obtain a decent Hall voltage. That is why this experiment would be much easier to do with a semiconductor, most of which have a much lower value of n.

Now, to determine the sign, notice that if the charge carriers were negative, the voltmeter as connected in Figure 11.9b would read negative. If instead the majority charge carriers are positive, the magnetic field will cause them to drift to the right on the strip. This is consistent with a positive voltmeter reading, and therefore we conclude that the majority charge carriers for zinc are positive. This is a somewhat unusual fact, though zinc is not unique—the majority carriers in Cd and Be are also positive. For most metals the charge carriers are negative.

Question 11.3: A very thin rectangular strip of zinc has been deposited on ...

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