Question

Figure 26-28wire section 1 of diameter D1 = 6.00R and wire section 2 of diameter D2 = 4.00R, connected by a tapered section. The wire is copper and carries a current. Assume that the current is uniformly distributed across any cross-sectional area through the wire's width. The electric potential change V along the length L = 2.50 shown in section 2 is 15.0 μV. The number of charge carriers per unit volume is 9.00 × 1028 m-3. What is the drift speed of the conduction electrons in section 1?

Answer 1

Given that diam eter \(D_{1}=6 R\)

diameter \(D_{2}=4 R\)

Length \(L=2.50 \mathrm{~m}\)

potential change \(V=15 \times 10^{-6} \mathrm{~V}\)

the number of charge carriers per unit volume is

\(n=9 \times 10^{20} \mathrm{~m}^{-3}\)

Current density of copper is \(\rho=1.69 \times 10^{-8} \Omega \mathrm{m}\)

The electric Field E is given by

\(E=\frac{V}{L}=\frac{15 \times 10^{-6}}{2.5}=6 \times 10^{-6} \mathrm{~V} / \mathrm{m}\)

hence the current density in section 2 is

\(J_{2}=\frac{E}{\rho}=\frac{6 \times 10^{-6}}{1.69 \times 10^{-8}}=355 \mathrm{~A} / \mathrm{m}^{2}\)

If \(J_{1}\) is the current density in section 1

then using conservation of electric current

\(J_{1} A_{1}=J_{2} A_{2}\)

\(J_{1}\left(\pi(3 R)^{2}\right)=J_{2}\left(\pi(2 R)^{2}\right)\)

\(9 J_{1} \pi R^{2}=4 J_{2} \pi R^{2}\)

\(J_{1}=\frac{4}{9} J_{2}\)

\(\quad=\frac{4}{9}(355)=157.77 \mathrm{~A} / \mathrm{m}^{2}\)

Now the drift speed on section 1 is

\(v_{d}=\frac{J_{1}}{n e}\)

\(=\frac{157.77}{\left(9 \times 10^{28}\right)\left(1.6 \times 10^{-19}\right)}\)

\(=10.956 \times 10^{-9} \mathrm{~m} / \mathrm{s}\)