The figure shows wire section 1 of diameter D_1 = 6.50R and wire section 2 of di

Question

The figure shows wire section 1 of diameter D_1 = 6.50R and wire section 2 of diameter D_2 = 4.00R, connected by a tapered section. The wire is copper and carries a current. Assume that the current is uniformly distributed across sectional area through the wire's width. The electric potential change V along the length L = 1.90 m shown in section 2 is 13.0 mu V. The number of charge carries per unit volume is 8.30 times 10^28 m^-3. What is the drift speed of the conduction electrons in section 1?

Explanation / Answer

driift speed of electrons is given as under
u = j/nq
where j is curent density
n is charge carrier density
q is charge on charge carrier

now, as current density is uniform everywhere we can say
j1 = current density at D1, j2 current density at D2
j1*D1^2 = j2*D2^2

also, drift velocity at D1 = u1
drift velocity at D2 = u2
u1*n*q *D1^2 = u2*n*q*D2^2
u1*D1^2 = u2*D2^2

also,
u = m*sigma*dV/rho *e*f*l

m = molecular mass of metal in kg = 63.546*1.6*10^-27 kg
sigma is conductivity of metal = 5.96*10^7
dV = voltage difference = 0.013 V
rho = density of material = 8960 kg/m^3
e = elementary charge = 1.6*10^-19 C
f = free electrons per atom = 2
l = 1.9 m
u2 = 1.446*10^-5 m/s
u1 = 1.446*10^-5*(4/6.5)^2 = 5.4762*10^-6 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.