Chapter 26, Problem 34 The figure wire section 1 of diameter D 5.50R and wire se

Question

Chapter 26, Problem 34 The figure wire section 1 of diameter D 5.50R and wire section 2 of diameter D 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 Valong the length L 1.50 m shown in section 2 is 15.0 uv. The number of charge carriers per unit volume is 8.10 x 10 m 3. What is the drift speed of the conduction electrons in section 1? (2) (1) Units Number The number of significant digits is set to 3; the tolerance is +/-2% Question Attempts: o of 10 used SAVE FOR LATER SUBMIT ANSWE

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 * (5.50R)^2 * v1 = pi * (4R)^2 *v2
=> v1 = v2 * 16/30.25

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

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