Question 14
(a)
The current I in a metal wire is
given by the expression
I = Anve.
State what is meant by
the symbols A and n.
[2]
(b)
The diameter of a wire XY varies linearly with distance along the
wire as shown in Fig. 7.1.
Fig. 7.1
There is a current I
in the wire. At end X of the wire, the diameter is d
and the average drift speed of the free electrons is vx. At end Y of the wire, the diameter is d
/ 2.
On Fig. 7.2, sketch a
graph to show the variation of the average drift speed with position
along the wire between
X and Y.
Fig. 7.2
[2]
[Total: 4]
Reference: Past Exam Paper – November 2018 Paper 22 Q7
Solution:
(a)
A: (cross-sectional) area (of wire)
n: number of free electrons per unit volume or number density of free
electrons
(b)
line drawn between (X, vx)
and (Y, 4vx)
line has increasing gradient
{I = Anve
Cross-sectional
area A = πd2 / 2
I = (Ï€d2 / 2) nve
As we move
from X to Y, the cross-sectional area (and hence, the diameter) decreases
linearly. So, we need to understand the relationship between drift velocity vx
and the diameter.
I = (Ï€d2 / 2) nve
v = 2I / πd2e
So, v ∝ 1 / d2
The drift velocity
v is inversely proportional to the square of the diameter. If the diameter
decreases, the drift velocity increases.
Initially at
X, drift velocity = vx. On the graph, we have the point (X, vx)
If the
diameter is halved, the the drift velocity is would increase by a factor of 4.
On the graph, we have the point (Y, 4vx).
Since v ∝ 1 / d2, the graph of
drift velocity against position (diameter) would resemble the reflection of the
graph of y = 1/x2 about the x-axis (since here, from X to Y, the
diameter is decreasing).}
Hi. I have a doubt in 9702_s19_qp_23 Q6. Any help would be thankful
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