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A potential difference of 3.25 nV is set up across a 2.19 cm length of copper wire that has a radius of 2.22 mm. How much charge drifts through a cross-section in 3.23 ms? Assume that the resistivity of copper is [tex]1.69 \times 10^{-8} \Omega \cdot m[/tex].

Answer :

Approximately [tex]\( 4.615 \times 10^{-26} \, \text{C} \)[/tex] of charge drifts through a cross-section of the wire in [tex]\( 3.23 \, \text{ms} \)[/tex].

To solve this problem, we can use the formula for drift velocity in a conductor and then calculate the charge that drifts through a cross-section of the wire in a given time interval.

The formula for drift velocity [tex](\( v_d \))[/tex] in a conductor is given by:

  • [tex]\[ v_d = \frac{I}{n \cdot A \cdot q} \][/tex]

Where:

  • - [tex]\( v_d \)[/tex] is the drift velocity

  • - [tex]\( I \)[/tex] is the current

  • - [tex]\( n \)[/tex] is the number density of charge carriers

  • - [tex]\( A \)[/tex] is the cross-sectional area of the conductor

  • - \[tex]( q \)[/tex] is the charge of a single charge carrier

Given that the potential difference [tex](\( V \))[/tex] across the wire is 3.25 nV, the length [tex](\( L \))[/tex] of the wire is 2.19 cm, and the radius [tex](\( r \))[/tex] of the wire is 2.22 mm, we can calculate the current [tex](\( I \))[/tex] using Ohm's law:

  • [tex]\[ V = I \cdot R \][/tex]

Where [tex]\( R \)[/tex] is the resistance of the wire, given by:

  • [tex]\[ R = \rho \cdot \frac{L}{A} \][/tex]

Given that the resistivity of copper [tex](\( \rho \))[/tex] is [tex]\( 1.69 \times 10^{-8} \Omega \cdot m \)[/tex], we can find the resistance [tex]\( R \)[/tex].

Once we have the current, we can find the drift velocity using the formula above. Finally, we can calculate the charge that drifts through a cross-section of the wire in a given time interval using the equation:

  • [tex]\[ Q = n \cdot A \cdot q \cdot v_d \cdot t \][/tex]

Given that [tex]\( t = 3.23 \)[/tex] ms, we will need to convert it to seconds before calculation.

Let's plug in the given values and solve step by step:

1. **Calculate the resistance [tex]\( R \)[/tex]**:

  • [tex]\[ R = \rho \cdot \frac{L}{A} \][/tex]

  • [tex]\[ R = (1.69 \times 10^{-8} \, \Omega \cdot m) \cdot \frac{0.0219 \, \text{m}}{\pi \times (0.00222 \, \text{m})^2} \][/tex]

  • [tex]\[ R ≈ (1.69 \times 10^{-8} \, \Omega \cdot m) \cdot \frac{0.0219 \, \text{m}}{3.1416 \times (4.932 \times 10^{-6} \, \text{m}^2)} \][/tex]

  • [tex]\[ R ≈ (1.69 \times 10^{-8} \, \Omega \cdot m) \cdot \frac{0.0219 \, \text{m}}{1.547 \times 10^{-5} \, \text{m}^2} \][/tex]

  • [tex]\[ R ≈ (1.69 \times 10^{-8} \, \Omega \cdot m) \cdot 1416.1 \, \Omega \][/tex]

  • [tex]\[ R ≈ 0.0239 \, \Omega \][/tex]

2. **Calculate the current [tex]\( I \)[/tex]** using Ohm's law**:

  • [tex]\[ V = I \cdot R \][/tex]

  • [tex]\[ I = \frac{V}{R} \][/tex]

  • [tex]\[ I = \frac{3.25 \times 10^{-9} \, \text{V}}{0.0239 \, \Omega} \][/tex]

  • [tex]\[ I ≈ 1.36 \times 10^{-7} \, \text{A} \][/tex]

3. **Calculate the drift velocity [tex]\( v_d \)[/tex]**:

  • [tex]\[ v_d = \frac{I}{n \cdot A \cdot q} \][/tex]

We need to find [tex]\( n \cdot A \)[/tex], the charge carriers per unit volume, which can be obtained from the volume of the wire's cross-section:

  • [tex]\[ n \cdot A = \frac{1}{V_c} \][/tex]

  • Where [tex]\( V_c \)[/tex] is the volume of the wire's cross-section.

  • [tex]\[ V_c = \pi \cdot r^2 \cdot L \][/tex]

  • [tex]\[ V_c = \pi \cdot (0.00222 \, \text{m})^2 \cdot 0.0219 \, \text{m} \][/tex]

  • [tex]\[ V_c ≈ 3.779 \times 10^{-8} \, \text{m}^3 \][/tex]

  • [tex]\[ n \cdot A = \frac{1}{3.779 \times 10^{-8} \, \text{m}^3} \][/tex]

Now, let's find \( v_d \):

  • [tex]\[ v_d = \frac{I}{n \cdot A \cdot q} \][/tex]

  • [tex]\[ v_d = \frac{1.36 \times 10^{-7} \, \text{A}}{\frac{1}{3.779 \times 10^{-8} \, \text{m}^3} \cdot q} \][/tex]

We need to find [tex]\( q \)[/tex], which is the charge of a single charge carrier. For copper, this is typically the charge of an electron [tex](\( e \))[/tex].

  • [tex]\[ q = e = 1.6 \times 10^{-19} \, \text{C} \][/tex]

Now we can calculate [tex]\( v_d \)[/tex]:

  • [tex]\[ v_d = \frac{1.36 \times 10^{-7} \, \text{A}}{\frac{1}{3.779 \times 10^{-8} \, \text{m}^3} \cdot 1.6 \times 10^{-19} \, \text{C}} \][/tex]

  • [tex]\[ v_d = \frac{1.36 \times 10^{-7} \, \text{A}}{6.0464 \times 10^{-27} \, \text{C/m}^3} \][/tex]

  • [tex]\[ v_d ≈ 2.25 \times 10^{-20} \, \text{m/s} \][/tex]

4. **Calculate the charge that drifts through a cross-section in 3.23 ms**:

  • [tex]\[ Q = n \cdot A \cdot q \cdot v_d \cdot t \][/tex]

We need to convert [tex]\( t \)[/tex] to seconds:

  • [tex]\[ t = 3.23 \times 10^{-3} \, \text{s} \][/tex]

Now, let's find [tex]\( Q \)[/tex]:

  • [tex]\[ Q = (3.779 \times 10^{-8} \, \text{m}^3) \cdot (1.6 \times 10^{-19} \, \text{C}) \cdot (2.25 \times 10^{-20} \, \text{m/s}) \cdot (3.23 \times 10^{-3} \, \text{s}) \][/tex]

  • [tex]\[ Q ≈ 4.615 \times 10^{-26} \, \text{C} \][/tex]

So, approximately [tex]\( 4.615 \times 10^{-26} \, \text{C} \)[/tex] of charge drifts through a cross-section of the wire in [tex]\( 3.23 \, \text{ms} \)[/tex].

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