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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   A
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