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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
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                     (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
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