Ch 28 current and conductivity
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Ch – 28 Current and Conductivity . Current: Motion of charge through a conductor. How do we know there is a current?. Charge carriers.

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Ch – 28 Current and Conductivity

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Ch 28 current and conductivity

Ch – 28 Current and Conductivity


Current motion of charge through a conductor

Current: Motion of charge through a conductor


How do we know there is a current

How do we know there is a current?


Charge carriers

Charge carriers

  • 18th century: Franklin and others developed the single fluid theory of electricity. Electricity flows from a body with an excess of charge (positive) to one with a deficit of charge (negative).

  • Theories of electricity advanced with the assumption of a positive charge carrier.

  • 19th century: Thompson and others suggested that negatively-charged electrons were the charge carriers, in a conductor.

  • Confirmed by Tolson and Stewart in 1916

  • Most engineering applications still assume positive charge carriers (aka electron holes)


Electron current aka actual current

Electron Current (aka actual current)

  • Sea of electrons move randomly but net motion=0 when conductor is in equilibrium

  • a force due to the presence of an external E field will cause the sea of electrons to move with vd - drift velocity

Moving sea of electrons


Electron current

Electron Current

  • The number of electrons per second that pass through a cross sectional area of wire or other conductor:

    i = Ne/∆t


Electron current cont d

Electron current, cont’d

Ne (number of electrons) = i ∆t

Ne = nV where V is the volume of the wire (A ∆x) and n is the conduction electron density (on the order of 1028 electrons per m3)

∆x = vd ∆t, therefore:

Ne = nAvd ∆t

Ne/∆t = i = nAvd


Stop to think

Stop to think

  • These four wires are made of the same metal. Rank, in order, from largest to smallest, the electron currents ia to id


Answer c b a d

Answer: c,b,a,d

i is proportional to r2vd. Changing r has more influence than changing v by the same amount


Numerical problem

Numerical Problem

1.0 x 1016 electrons flow through a x-section of silver wire in 320 μs with a drift velocity of 8.0 x 10 -4 m/s. What is the diameter of the wire?


Answer

Answer

9.26 x 10-4 m


Conservation of current

Conservation of Current

  • The drift velocity of electrons is the same throughout the wire

  • The electrons themselves can’t go anywhere while traveling through the wire

  • Therefore the current going in is equal to the current coming out


Creating a current

Creating a current

An electron current is a non-equilibrium motion of charges sustained by an electric field


E field in a wire

E field in a wire


E field in a wire1

E field in a wire

  • On-axis field for charged ring

    • points away from positive charge, towards ring for negative charge

    • is proportional to the charge on the ring

    • decreases with distance from the ring


Stop to think 2 page 886 in text

Stop to think# 2, page 886 in text


Answer1

Answer

d>b>e>a=c

E depends on the difference in the charge on the two rings. The E fields of a and c are zero. The difference is the greatest for the rings of d.


Problem

Problem

  • What is the surface charge density of a 1.0 mm-diameter wire with 1000 excess electrons per cm of length?


Answer2

Answer

ή = 5.1 x 10-12 C/m2


Conventional current

Conventional Current

  • The rate in coulombs per second, at which charge moves in the direction of E

    • For constant current I = ∆Q/∆t

    • For changing current I = dQ/dt

  • Current direction from positive terminal to negative terminal, opposite the direction of electrons in a metal

  • I = ∆Q/∆t = -(eNe/ ∆t) = -ei (sign for direction)


Problem constant current

Problem- constant current

In an ionic solution, 5.0 x 1015 positive ions with charge +2e pass to the right, while 6.0 x 1015 negative ions with charge –e pass to the left. What is the current in the solution and what is the direction of the E field?


Answer constant current

Answer –constant current

2.56 mA (milliamps).

E field is to the right


Problem changing current

Problem – changing current

The current in a wire at time t is given by the expression: I = (2.0 A)e-t/(2.0μs) where t is in μs and t>=0.

  • Graph I vs t for 0<=t<=10 μs (2 μs intervals)

  • Find an expression for the total amount of charge that is entering the wire at time t. Q=0C at t=0 μs.

  • Graph Q vs t for 0<=t<=10 μs (2 μs intervals).


Answer changing current

Answer – changing current

  • See top graph

  • Q = (4.0 μC)[1- e-t/(2.0μs)]

  • See bottom graph


Current density in a wire

Current Density in a Wire

  • I = ei = nevd A

  • Define current density as :

    J = I/A = nevd (A/m2)

  • This quantity is not the same as surface charge density, which implies electrostatic conditions (no moving charge)


Current density conceptual questions

Current Density Conceptual Questions

The current in wire is doubled. By what factor do the following change?

  • Current density

  • Conduction-electron density

  • Electron drift speed


Current density conceptual answers

Current Density Conceptual Answers

  • J increases by a factor of 2 (J = I/A)

  • n remains the same (property of the metal)

  • vd increases by a factor of 2 (J = nevd) and e is the charge on the electron, which is constant.


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