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There are 8.4 x 1022 free electrons per cm incurrent in the wire is 0.21 A(e = 1.6 x 10-19 C). Then the drifts velocity ofelectrons in a copper wire of 1 mm2 cross section,will be :-(1) 2.12 x 10-6 m/s (2) 0.78 x 10-5 m/s(3) 1.56 x 10-6 m/s (4) none of theseplss give the whole calculation |
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Answer» Answer: 4 Step-by-step explanation: current, i equal to 1A Area of cross -section A = 1 millimeters square = 1 × 10 Ka power -6 meter square length of the conductor = l Also Mass of COPPER wire = volume into density m =A×L×P=》m=Al×9000 we know that the NUMBER of ATOMS in molecular mass, M=NA therefore No. of atoms in mass, m,N= (NA/M )m where NA is known as Avogadaro no. and is equal to 6×10 Ka power 23 =》N=(NA/M)m=》N=(NA/M)×A×l×9000 Also, it is GIVEN that No. of free electron =No. of atom Let n be the no. of free electron per unit volume n=No. of electron/volume =NA×A×l×9000/M×A×like NA×9000/m=6×10 Ka power 23/63.5× 10 Ka power -23 therefore i=VdnAe=》1/6×10ka power 23×9000/63.5×10ka power-3 ×10ka power-6×1.6×10 Ka power -19=63.5×10ka power-3/6×10 Ka power 23×9000×10ka power -6×1.6×10ka power -19=63.5×10ka power -3/6×10 Ka power 26×9×10ka power -6×1.6×10ka power -19=63.5×10ka power -3/6×9×16=0.073×10 Ka power -3 m/s=0.073 m/ s so the answer is 4 |
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