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Physics Semester 2  STPM  Chapter 14 Electric Current
                       (ii)  In 1.0 m  of copper, 8.466 × 10  atoms of copper contribute
                                  3
                                                    28
                           8.34 × 10  free electrons.
                                   28
                           8.466 × 10  atoms ⎯→ 8.34 × 10  free electrons
                                                      28
                                   28
                                              8.34 × 10 28
                                \ 1 atom ⎯→
                                              8.466 × 10 28
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                                             = 0.985 electrons
                           \ Each copper atom contributes 0.985 free electron.

                 14.3 Current Density

                 Section  A   Multiple-choice Questions

                Question 1
                  A  current  of  3.0  A flows  in  an  aluminium  wire  of  cross-sectional  area
                  2.5 × 10  m . What is the current density in the wire?
                            2
                         –6
                  A  1.2 × 10  A m               C  7.5 × 10  A m –2
                                –2
                                                          6
                           6
                                –2
                           6
                                                          6
                  B  1.5 × 10  A m               D  8.3 × 10  A m –2
                Answer: A
                   Current density, J =   I   =   3                                   Semester
                                    A    2.5 × 10 –6
                                               6
                                       = 1.2 × 10  A m
                                                    –2
                                                                                      2
                Question 2
                  There are 6.23 × 10  free electrons per cm  in a wire. The average drift velocity
                                                    3
                                  24
                  of the electrons is 1.1 × 10  m s . Calculate the current density in the wire.
                                            –1
                                       –3
                  A  1.0 × 10  A m               C  2.1 × 10  A m –2
                                –2
                                                          9
                           3
                  B  6.9 × 10  A m –2            D  1.1 × 10  A m –2
                                                          9
                           6
                Answer: D
                   J  = nve
                        6.23 × 10
                     =  1  1 × 10  m 24 3 2 (1.1 × 10  m s ) (1.6 × 10  C)
                                          –3
                                                        –19
                                              –1
                             –6
                                                          Exam Tips
                     = 1.1 × 10  C s  m –2                Exam Tips
                                 –1
                             9
                                                              I
                     = 1.1 × 10  A m –2                    J =   =   nAve   = nve
                             9
                                                              A    A
                                                           n = density of electrons
                                                           v = drift velocity
                                                227
         014 STPM Q&A Physics T2.indd   227                                  17/02/2022   10:12 AM
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