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Physics Semester 2  STPM  Chapter 14 Electric Current

                 Section  B   Structured Questions
                Question 5
                  A current of 1.1 A flows in a copper wire of diameter 1.2 mm. The density of
                  copper is 8.93 × 10  kg m  and the relative atomic mass of copper is 63.5.
                                 3
                                       –3
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                  Assuming that each copper atom contributes one free electron, calculate
                                                    3
                  (a)  the number of free electrons in 1.0 m  of wire,
                  (b)  the current density in the wire,
                  (c)  the drift velocity of the electrons in the wire.
                Answer:
                   (a)  Density =   mass                  Exam Tips
                                volume                    Exam Tips
                                  N   × molar mass         Number of
                            ρ =   N A                        atoms   =   Mass
                                       V                    Avogdro   Molar mass
                           N   =   ρN A                     number
                           V     M                                 N   =   m
                                                                        M
                                                 23
                                        3
                              =  (8.93 × 10 )(6.02 × 10 )          N A
                                     63.5 × 10 –3
                              = 8.466 × 10 28
                       \ Number of the electrons per m  = 8.466 × 10 28               Semester
                                                  3
                   (b)  Cross-sectional area, A =   πd 2
                                              4
                                                     –3 2
                                          =  π(1.2 × 10 )                             2
                                                  4
                                          = 1.131 × 10  m 2
                                                    –6
                       Current density, J  =   I
                                         A
                                      =     1.1
                                        1.131 × 10 –6
                                      = 9.726 × 10  A m –2
                                                5
                   (c)   J  = nve
                       v  =   J
                           ne
                         =       9.726 × 10 5
                           (8.466 × 10 )(1.6 × 10 )
                                     28
                                              –19
                         = 7.18 × 10  m s –1
                                  –5


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         014 STPM Q&A Physics T2.indd   229                                  17/02/2022   10:12 AM
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