A copper wire with a 3mm diameter is connected across abattery. At 20*C a 6A cur
ID: 1670959 • Letter: A
Question
A copper wire with a 3mm diameter is connected across abattery. At 20*C a 6A current flows through the wire. At 20*C:copper= 1.7x10-8 ohms-meter andcopper= .0039/C*. a) What is the electric field inside the wire? b) If the temperature is increased to 50*C then what is thecurrent through the wire, assuming the battery's voltage remainsconstant? Will rate LIFESAVER!! Thanks. A copper wire with a 3mm diameter is connected across abattery. At 20*C a 6A current flows through the wire. At 20*C:copper= 1.7x10-8 ohms-meter andcopper= .0039/C*. a) What is the electric field inside the wire? b) If the temperature is increased to 50*C then what is thecurrent through the wire, assuming the battery's voltage remainsconstant? Will rate LIFESAVER!! Thanks.Explanation / Answer
a. Currentdensity J = I/ A Area ofcross section of thewire A = *r2 = 3.14* (3.00 * 10-3/ 2)2 = 7.065* 10-6 m2 J = 6/ 7.065 * 10-6 = 8.49* 105 A/m2 electric field insidewire E = J* = 8.49* 105 * 1.7 * 10-8 = 1.44* 10-2 V/m = 1.44* 10-2 V/m b. Resistivity attemperature T2 T2 = T1* {1 + *(T2 - T1)} 50 = 1.7* 10-8 * {1 + 0.0039 * (50 - 20)} = 1.899*10-8 -m Initial resistance R20 = 20* L / A Initialcurrent I20 = V/ R20 = V* A / 20 * L Finlacurrent I50 = V* A / 50 * L Dividing I50 byI20 I50 /I20 = ( V* A / 50 * L) / ( V * A / 20 *L) = 20/ 50 I50 = I20* (1.7 * 10-8 / 1.899 * 10-8) = 6* 1.7 / 1.899 = 5.37 A I50 = I20* (1.7 * 10-8 / 1.899 * 10-8) = 6* 1.7 / 1.899 = 5.37 ARelated Questions
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