Q#13 (Past Exam Paper – November 2018 Paper 23 Q6)
(a) Define the coulomb. [1]
(b) An electric current is a flow of charge carriers.
In the following list, underline the possible charges for a charge carrier.
8.0 × 10-19 C 4.0 × 10-19 C 1.6 × 10-19 C 1.6 × 10-20 C [1]
(c) The diameter of a wire ST varies linearly with distance along the wire as shown in Fig. 6.1.
Fig. 6.1
There is a current I in the wire. At end S of the wire, the diameter is d and the average drift speed of the free electrons is vs. At end T of the wire, the diameter is 2d.
On Fig. 6.2, sketch a graph to show the variation of the average drift speed with position
along the wire between S and T.
Fig. 6.2
[2]
[Total: 4]
Solution:
(a) The coulomb is the charge that passes a point in a circuit when there is a current of 1 A for 1 s.
(b) 8.0 × 10-19 C and 1.6 × 10-19 C both underlined (and no others underlined)
{The charge can only have an integer value times the elementary charge (charge of electron = 1.6×10-19 C)
8.0 × 10-19 C / 1.6×10-19 C = 5}
(c)
line drawn between (S, 1.00vs) and (T, 0.25vs)
line with decreasing magnitude of gradient
{At end S of the wire, the average drift speed of the free electrons = vs. This corresponds to point (S, 1.00vs) on the graph.
Electric current: I = Anvq
Average drift speed v = I / Anq
The current I flowing in the wire, the number density n of electrons in the wire and the charge q of an electron are constants.
So, v is inversely proportional to cross-sectional area A.
Area A = πd2 / 4
At end T of the wire, the diameter is doubled (= 2d). This means that the area A is 4 times greater.
Since v is inversely proportional to area A, when the area at T is 4 times that at S, the average drift speed at T would be ¼ (= 0.25) that at S. This corresponds to point (T, 0.25vs) on the graph.
As we move from S to T, the diameter varies linearly with the position (the diameter increases linearly).
We have seen that v ∝ 1/A and A ∝ d2
So, v ∝ 1/d2
The average drift velocity is inversely proportional to the square of the diameter.
This resembles the graph of y = 1/x2 which has a decreasing gradient.}
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