The circuit is in the steady state. The number of electrons passing location B e

Question

The circuit is in the steady state. The number of electrons passing location B each second is the same as the number of electrons passing location D each second. The drift speed of electrons passing location D is greater than the drift speed of electrons passing location B.
The radius of the thin wire is 0.15 mm, and the radius of the thick wire is 0.375 mm. There are 4e+28 mobile electrons per cubic meter of this material, and the electron mobility is 0.0006 (m/s)/(V/m). If 5e+18 electrons pass location D each second, how many electrons pass location B each second? (Note at location D the wire is thin and at B the wire is thick.)
A) _______ electrons per second

What is the magnitude of the electric field at location B ?
B) ________ V/m

Explanation / Answer

A)

The current must be constant along a wire (KCL).

Current = charge flowing / time

Charge that enters a wire segment must be constant along a wire

Charge = charge of an electron * number of electrons

So the same amount of electrons enters each segment of wire in any given time.

If the two wires are connected in series, they act as a single wire of varying thickness.

If 5e18 electrons per second pass through a cross-section of one wire, 5e18 electrons per second must pass through any cross-section of the other wire.

At D,

Cross section area = pi * r * r = 0.07065 mm^2

Through 0.07065 mm^2 , numbers of electron passing is 5e+18,

Therefore for 1 mm^2 , number of electron passing is 5e+18 / 0.07065

Therefore at B,

Area of cross section = pi * R * R = 0.4416 mm^2

Therefore number of electron passing = 0.4416 * (  5e+18 / 0.07065 )

= 31.25e+18 ANS

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