You have a 2 mm diameter copper wire connected to a 1 mm diameter gold wire as s

Question

You have a 2 mm diameter copper wire connected to a 1 mm diameter gold wire as shown below. A 15 A current is put into the copper wire: a. What will the current I be in the gold wire? b. What is the current density J in both wires? c. What is the electric field E in each wire assume no edge effects) d. What is the electron drift velocity v_d in each wire? e. What is the electron current i_e and what direction does it flow in each wire? f. How would you need to change the gold wire if you wanted the electric field strength to be the same in both wires?

Explanation / Answer

FOR POWER I HAVE WRITTEN 10e -9 if the power of 10 is -9 so plz see carefully

a) current wil be same in both the wires i.e.15A

b)Current density I/A so for copper 15/3.14×104=4.8×104A/m2

Current density for gold wil be 15/((0.5) 2 ×3.14) ×104=19.11×104A/m2

c) ELECTRIC FIELD IN COPPER = RESISTIVITY×CURRENT DENSITY=1.7×10e-8×4.8×10e4 =8.16×10e-4 V/m

Electric field in gold =46.62 ×10e-4V/m using resistivity of gold is 2.44×10e-8ohm m

d) drift velocity in copper wil be I /neA =15/8.4×10e 28 ×1.6× 10e-19 ×1×3.14×10e-4 =0.36×10e -5 m/s

drift velocity in gold will be n =5.89×10e28 so v= 0.51×10e-5m/s

we can cal n by the formula density×avagadro no /atomic mass density of gold is 19300kg/m^3

atomic mass 196.96 u

e) electron cirrent wil be 10A but in opposite direction to flow of current

d) we have to increase diameter of gold wire to have same electric field strength

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