Question

A copper wire of diameter 1 mm and length 30 m is connected across a battery of
2V. Calculate the current density in the wire and drift velocity of the electrons. The
resistivity of copper is 1.72 × 10−8 Wm and n = 8.0 × 1028 electrons m−3.

Answer 1

Answer on Question 69084, Physics, Electromagnetism

Question:

A copper wire of diameter $1 \, \text{mm}$ and length $30 \, \text{m}$ is connected across a battery of $2 \, \text{V}$. Calculate the current density in the wire and drift velocity of the electrons. The resistivity of copper is $1.72 \cdot 10^{-8} \, \Omega \cdot \text{m}$ and $n = 8.0 \cdot 10^{28} \, \text{electrons/m}^3$.

Solution:

a) Let's first find the resistance of the copper wire from the formula:

$$R = \rho \frac{l}{A},$$

here, $\rho$ is the resistivity of the copper wire, $l$ is the length of the wire, $A = \frac{\pi d^2}{4}$ is the cross-sectional area of the wire and $d$ is the diameter of the wire.

Then, we get:

$$R = \rho \frac{l}{A} = \rho \frac{4l}{\pi d^2} = 1.72 \cdot 10^{-8} \, \Omega \cdot \text{m} \cdot \frac{4 \cdot 30 \, \text{m}}{\pi \cdot (1.0 \cdot 10^{-3} \, \text{m})^2} = 0.65 \, \Omega.$$

Then, from the Ohm's law we can find the current flowing through the copper wire:

$$I = \frac{V}{R} = \frac{2.0 \, \text{V}}{0.65 \, \Omega} = 3.1 \, \text{A}.$$

Finally, we can find the current density in the wire:

$$J = \frac{I}{A} = \frac{4I}{\pi d^2} = \frac{4 \cdot 3.1 \, \text{A}}{\pi \cdot (1.0 \cdot 10^{-3} \, \text{m})^2} = 3.95 \cdot 10^6 \, \frac{\text{A}}{\text{m}^2}.$$

b) We can find the drift velocity of the electrons from the formula:

$$v = \frac{I}{n A q},$$

here, $I$ is the current flowing through the wire, $n$ is the number of free electrons per unit volume of the copper wire, $A$ is the cross-sectional area of the wire and $q$ is the charge on each electron.

Then, we get:

$$v = \frac {I}{n A q} = \frac {4 I}{n \pi d ^ {2} q} = \frac {4 \cdot 3 . 1 A}{8 . 0 \cdot 1 0 ^ {2 8} \frac {\text {electrons}}{m ^ {3}} \cdot \pi \cdot (1 . 0 \cdot 1 0 ^ {- 3} m) ^ {2} \cdot 1 . 6 \cdot 1 0 ^ {- 1 9} C} = 3. 0 8 \cdot 1 0 ^ {- 4} \frac {m}{s}.$$

Answer:

a) $J = 3.95 \cdot 10^{6} \frac{A}{m^{2}}$ .

b) $v = 3.08 \cdot 10^{-4} \frac{m}{s}$ .

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