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Wednesday, January 21, 2015

Charge and Current

Charge and Current

•The charge carriers are usually electrons; q = e=  -1.60 x 10-19 C; but be careful when a current of ions exists.

The value of n for copper is 8.0 x 1028 m3.

  1. Calculate the current in a copper wire with area of cross section 2.0 x 10-5 m2 when electrons drift through it with a mean speed of 0.80 mms-1
I = nAve = (8.0 x 1028) (2.0 x 10-5) (0.80 x 10-3) (-1.60 x 10-19) = -205 A

  1. In a silicon transistor with area of cross section 3.8 x 10-6 m2 there is a d.c. current of 200 mA. The current is a flow of electrons and the number density of free electrons for silicon is 8.3 x 1023 m3. Determine the mean drift speed of the electrons.
I = nAve
rearranging v = I/nAe
v = (200 x 10-3) / (8.3 x 1023) (3.8 x10-6) (-1.60 x 10-19)
v = -0.40 ms-1

  1. The beam of electrons in a cathode-ray tube contains electrons travelling at 8.4 x 106 ms-1 and the current through the tube is 2.8 μA.
    (a) How many electrons are emitted per second from the cathode of the tube?
Q= It = 2.8 x 10-6 x 1 second = 2.8 x 10-6 C
Number of electrons = total charge / e = 2.8 x 10-6 /1.6x 10-19  
= 1.75 x 1013

(b) What is the number of electrons per unit length of the beam?
In 1 s an electron will travel 8.4 x 106 m. this means that 1.75 x1013 will be spread over a length of 8.4 x 106 m. So 1 m will contain 1.75 x1013 electrons ÷ 8.4 x 106 m = 2.08 x 106 electrons

  1. A direct current of 3.0 A through a copper wire reaches a place where the area of cross section of the wire changes from 2.0 x 10-6 m2 to 0.090 x l0-6 m2. By what factor does the drift speed of the delocalised  electrons increase as they move from the wide to the narrow section? Suggest how this indicates that a damaged wire will overheat.
The same current flows through the thick wire as the thin wire, (Kirchoff’s first law)

For the thick part of the wire
I = nAve
rearranging v = I/nAe
v = 3 / (8.0 x 1028) (2 x10-6) (-1.60 x 10-19)
v = -1.17 x 10-4 ms-1
For the thin part of the wire
I = nAve
rearranging v = I/nAe
v = 3 / (8.0 x 1028) (0.090 x10-6) (-1.60 x 10-19)
v = -2.60 x 10-3 ms-1

Ratio = -2.60 x 10-3 ms-1 / -1.17 x 10-4 ms-1 = 22.2

Alternatively
I = nAwvwe = nAnvne
As n and v are constants then Awvw = Anvn
rearranging Aw/An=vn/vw      Aw = 2.0 x 10-6 m2 and An = 0.090 x l0-6 m2
so Aw/An = 2.0 x 10-6/ 0.090 x l0-6 = vn/vw = 22.2

This increase in drift speed will cause more collisions per second between electrons and the crystal lattice transferring more energy to heat

  1. Explain why a light comes on almost immediately when switched on, although the drift speed of the electrons in the wires to the light is so small.
When the circuit is made the electric field propagates through the wire at close to the speed of light. This imposes a net direction from negative to positive on the hitherto random motion of the delocalised electrons



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