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Physics Semester 2 STPM Chapter 14 Electric Current
Question 6
(a) Define electric current density and conductivity.
(b) (i) When a potential difference, V is applied across the ends of a metallic
wire of length l, cross-sectional area A and resistance R, a current I
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flows through the wire. Deduce an expression to show the relationship
between current density and conductivity.
(ii) There are 5.0 × 10 free electrons in one cubic metre of the wire. If
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the potential gradient applied across the wire is 2.0 V m and the drift
velocity of the electrons is 3.91 × 10 m s , calculate the conductivity
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of the wire.
Answer:
(a) Electric current density, J = electric current, I
cross-sectional area, A
Conductivity, σ = 1
resistivity, ρ
Exam Tips
Exam Tips
(b) (i) Electric field strength E at any point dV V
in the wire, E = – dx or l
E = V ⇒ V = El E = electric field
l strength
dV
Current density, J = I ⇒ I = JA dx = potential
A gradient
Semester
2 Resistance of wire, R = ρ l
A Exam Tips
= l ⇐ σ = 1 Exam Tips
σA ρ To find unit of σ:
Ohm’s Law: V = IR ρ = RA
l
l
El = JA 1 2 ⇒ (Ω)(m ) = Ωm
2
σA
\ J = σE 1 m
σ =
ρ
–1
(ii) J = nve = Ω m –1
σE = nve
σ(2) = (5 × 10 )(3.91 × 10 )(1.6 × 10 )
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7
σ = 1.56 × 10 Ω m –1
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014 STPM Q&A Physics T2.indd 230 17/02/2022 10:12 AM

