The figure below shows wire section 1 of diameter D 1 = 5.00R and wire section 2
ID: 2055104 • Letter: T
Question
The figure below shows wire section 1 of diameter D1 = 5.00R 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 = 1.70 m shown in section 2 is 11.0 V.
The number of charge carriers per unit volume is 8.10 × 1028 m-3.
What is the drift speed of the conduction electrons in section 1?
The figure below shows wire section 1 of diameter D1 = 5.00R 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 = 1.70 m shown in section 2 is 11.0 MuV. The number of charge carriers per unit volume is 8.10 1028 m^-3. What is the drift speed of the conduction electrons in section 1?Explanation / Answer
We know that i = n*A*v*Q where I is the electric current n is number of charged particles per unit volume (or charge carrier density) A is the cross-sectional area of the conductor v is the drift velocity Q is the charge on each particle. Given, i = constant => n*A*v*Q = constant => A1 *v1 = A2 * V2 => pi * (5R)^2 * v1 = pi * (4R)^2 *v2 => v1 = v2 * 16/25 Use I = V/R & I = v.A.n.Q, R = p.A/l to get the answer for v1. Take resistivity(p) value of copper as 1.68×10-8 We will get the answer very easily. If you want, I can solve further