Chapter 4: Q12E (page 842)

A copper wire has a square cross-section 2.3 mm on a side. The wire is 4.0 m long and carries a current of 3.6 A. The density of free electrons is $8.5 \times 10^{28} / m^{3}$ . Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

Short Answer

Expert verified

Current density in the wire is $J = 6.8 \times 10^{5} A / m^{2}$
Electric field in the wire is $E = 11.69 \times 10^{- 3} V / m$
Time required by the electron $τ = 22.22 hr$

Step by step solution

Step 1:

Given data:

_{$A = l^{2} = {(2.3 \times 10^{- 3} m)}^{2} = 5.3 \times 10^{- 6} m^{2} L = 4.0 m l = 3.6 A$}

Density of free electrons $= 8.5 \times 10^{28} / m^{3}$

(a) Current density in the wire is:

_{$J = \frac{I}{A} = \frac{3.6 A}{5.3 \times 10^{- 6} m^{2}} = 6.8 \times 10^{5} A / m^{2}$}

Therefore, the current density in the wire is $J = 6.8 \times 10^{5} A / m^{2}$ .

Step 2:

(b) Electric field in the wire is:

As the copper resistivity is $ρ = 1.72 \times 10^{- 8} Ω \cdot m$

So electric field is:

_{$E = ρ J = (1.72 \times 10^{- 8} Ω \cdot m) (6.8 \times 10^{5} A / m^{2}) = 0.01169 V / m = 11.69 \times 10^{- 3} V / m$}

Hence, the electric field in the wire is $E = 11.69 \times 10^{- 3} V / m$ .

Step 3:

From the equation,

_{$τ = \frac{L}{v_{d}}$}

Here $v_{d}$ is drift speed;

$v_{d} = \frac{J}{n |q|}$

So, the time required is:

$τ = \frac{L}{v_{d}} = \frac{L \cdot n |q|}{J} = \frac{(4.0 m) \times (8.5 \times 10^{28}) \times (|- 1.6 \times 10^{- 19} C|)}{6.8 \times 10^{5} A / m^{2}} = 80000 s = 1333.33 \min = 22.22 hr$

Therefore, the time required by the electron $τ = 22.22 hr$ .

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