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.
- 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
- In a silicon transistor with
area of cross section 3.8 x 10-6 m2 there is a
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
- 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
- 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
- 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
