Question

An 18 gauge copper wire (the size usually used for lamp cords), with a diameter of 1.02 mm,1.02 mm, carries a constant current of 1.67 A1.67 A to a 200 W lamp. The free-electron density in the wire is 8.5×10288.5×1028 per cubic meter. Find (a) the current density and (b) the drift speed.

Answer 1

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Answer 2

J = 2.044x10⁶ A/m²

v = 1.50x10⁻⁴ m/s

Explanation:

The current density (J) of the copper wire is giving by:

`J = \frac {I}{A}`

where I: electric current and A: cross-sectional area of the copper wire

The cross-sectional area of the copper wire can be calculated by:

`A = \frac {\pi d^(2)}{4} = \frac {\pi (1.02 \cdot 10^(-3) m)^(2)}{4} = 8.17 \cdot 10^(-07) m^(2)`

Substituting the calculated area in the equation (1) we have:

`J = \frac {1.67 A}{8.17 \cdot 10^(-7) m^(2)} = 2.044 \cdot 10^(6) \frac {A}{m^(2)}`

Hence, the current density is 2.044x10⁶ A/m².

To find the drift speed (v), we need to use the next equation:

`v = \frac {J}{n q}`

where n: the free-electron density, q: module of the charge of the electron

`v = \frac {2.044 \cdot 10^(6) \frac {A}{m^(2)}}{(8.5 \cdot 10^(28) {m^(-3)}) (1.6 \cdot 10^(-19) C)}`

`v = 1.50 \cdot 10^(-04) \frac {m}{s}`

So, the drift speed is 1.50x10⁻⁴ m/s.

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