Currents in solids
Some copper
fuse wire has a diameter of 0.22mm and is designed to carry currents of up to
5.0 A. If there are 1.0 x 1029 electrons per m3 of
copper, what is the mean drift speed of the electrons in the fuse wire when it
carries a current of 5.0A?
I
= nAve rearranging v = I/nAe
Area
= ∏r2 r = 0.22 x 10-3 /2 so ∏r2 = ∏(0.11 x 10-3)2
= 3.8 x 10-8 m2
So
v = 5.0/ (1.0 x 1029)( 3.8 x 10-8)(1.6 x 10-19)
= 8.2 x 10-3 ms-1
A wire
carrying an electric current will overheat if there is too large a current: the
accepted value for the maximum allowable current in a copper wire is 1.2 x 107
A per square metre of cross-section of the wire. If there are 1.0 x 1029
electrons per m3 of copper, calculate the mean drift speed of the
electrons in the wire when the current reaches this value.
I
= nAve rearranging v = I/nAe
So
v = 1.2 x 107
/ (1.0 x 1029)(1)(1.6 x 10-19) = 7.5 x 10 -4 ms-1 =
0.75 mms-1
Two copper
wires of diameter 2.00 mm and 1.00mm are joined end-to-end. What is the ratio
of the average drift speeds of the electrons in the two wires when a steady
current passes through them? In which wire are the electrons moving faster?
I
= nAve rearranging v = I/nAe
Area
= ∏r2 r
For
d = 1.00 mm r = 1 x 10-3 /2 so ∏r2 = ∏(0.5 x 10-3)2
= 7.85 x 10-7 m2
I
don’t know the current so I, but I do know it is the same in both so let’s say
it is 1A
So
v = 1/ (1.0 x 1029)( 7.85 x 10-7)(1.6 x 10-19)
= 7.9 x 10-5ms-1
For
d = 2.00 mm r = 2 x 10-3 /2 so ∏r2 = ∏(1 x 10-3)2
= 3.14 x 10-6 m2
I
don’t know the current so I, but I do know it is the same in both so let’s say
it is 1A
So
v = 1/ (1.0 x 1029 (3.14 x 10-6)(1.6 x 10-19)
= 1.99 x 10-5ms-1
Ratio
of 2mm/1mm = 1.99/7.9 = 0.25 = 1:4
Alternate
I = n1A1v1e1
= n2A2v2e2 As e and n are constant A1v1 =
A2v2 and v2/v1 = A1/A2
as A = ∏r2 = ∏(d/2)2 then A1 = ∏(1/2)2
and A2 =∏(2/2)2
so A1/A2 = ∏(1/2)2/∏(2/2)2 = 1/4
A copper
wire joins a car battery to one of the tail lamps and carries a current of
1.8A. The wire has a cross-sectional area of 1.0 mm2 (1x 10-6m2)and is 6.0 m long.
If there are 1.0 X 1029 electrons per m3 of copper,
calculate how long it takes an electron to travel along this length of wire.
I
= nAve rearranging v = I/nAe
So
v = 1.8 / (1.0 x 1029)(1 x 10-6)(1.6 x 10-19)
= 8.2 x 10-3 ms-1 = 1.125 x 10 -4 ms-1
v = s/t rearranging t = s/v = 6m / 1.125 x 10 -4
ms-1 = 5.33 x 104 s = 14.8 hrs
