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Physics Semester 2  STPM  Chapter 14 Electric Current
                Question 3

                  A conductor of length 2.0 m and cross-sectional area 5.0  × 10  m   carries
                                                                          2
                                                                       –6
                  a  current  of  6.0  A.  If  the  drift  velocity  of  the  conduction  electrons  is
                  1.5 × 10  m s , what is the number of electrons in the conductor?
                         –4
                             –1
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                  A  2.5 × 10                    C  4.2 × 10 23
                           23
                  B  2.9 × 10 23                 D  5.0 × 10 23
                Answer: D
                   I  = nAve
                      =   N Ave =  Nve
                       Al       l
                   N =   Il
                       ve
                     =         6(2)
                       (1.5 × 10 )(1.6 × 10 )
                              –4
                                        –19
                     = 5.0 × 10
                             23

                Question 4
                  A tungsten wire of length 1.00 m and cross-sectional area 2.01 × 10  m  carries
                                                                          2
                                                                       –6
                  a current of 5.00 A.
                  If the number of free electrons per unit volume is 4.3 × 10 , what is the time   Semester
                                                                 28
                  taken by an electron to travel from one end of the wire to the other end?
                  A  1.81 × 10  s                C  4.51 × 10  s
                            3
                                                           3
                  B  2.77 × 10  s                D  7.25 × 10  s                      2
                                                           3
                            3

                Answer: B
                   I  = nAve
                   v  =   I
                        nAe
                     =                5
                        (2.01 × 10 )(4.3 × 10 ) (1.6 × 10 )
                                         28
                                                   –19
                                –6
                     = 3.616 × 10  m s –1
                               –4
                   t  =   l
                        v
                     =      1
                        3.616 × 10 –4
                     = 2.77 × 10  s
                              3


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         014 STPM Q&A Physics T2.indd   223                                  17/02/2022   10:12 AM
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