Part $$ -A $$ 1.) A thin cylindrical wire carries a current of 1.67 Amperes and has a diameter of 1.02 mm. The resistivity of the material of the wire is $$ 1.72 imes 10^{\circ} \Omega \mathrm{m} $$. (a) What is the magnitude of the current density in the wire? (b) What is the magnitude of the electric field in the wire? (c) What is the potential difference between two points in the wire 50 mn apart? (d) What is the resistance of a 50 m length of this wire?

Question

Part $$ -A $$ 1.) A thin cylindrical wire carries a current of 1.67 Amperes and has a diameter of 1.02 mm. The resistivity of the material of the wire is $$ 1.72 	imes 10^{\circ} \Omega \mathrm{m} $$. (a) What is the magnitude of the current density in the wire? (b) What is the magnitude of the electric field in the wire? (c) What is the potential difference between two points in the wire 50 mn apart? (d) What is the resistance of a 50 m length of this wire?

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**(a) Calculation of the current density**

1. Given a cylindrical wire of diameter  
   $$
   d = 1.02\,\text{mm} = 1.02 \times 10^{-3}\,\text{m},
   $$
   the radius is  
   $$
   r = \frac{d}{2} = 0.51 \times 10^{-3}\,\text{m}.
   $$

2. The cross‐sectional area is  
   $$
   A = \pi r^2 = \pi (0.51 \times 10^{-3})^2.
   $$
   Compute  
   $$
   (0.51 \times 10^{-3})^2 = 0.2601 \times 10^{-6} = 2.601 \times 10^{-7}\,\text{m}^2.
   $$
   Thus,  
   $$
   A \approx \pi (2.601 \times 10^{-7}) \approx 8.174 \times 10^{-7}\,\text{m}^2.
   $$

3. With the current  
   $$
   I = 1.67\,\text{A},
   $$
   the current density is  
   $$
   J = \frac{I}{A} = \frac{1.67}{8.174 \times 10^{-7}} \approx 2.04 \times 10^{6}\,\text{A/m}^2.
   $$

---

**(b) Calculation of the electric field in the wire**

1. The relationship between the electric field and the current density is given by  
   $$
   E = \rho J,
   $$
   where the resistivity of the material is  
   $$
   \rho = 1.72 \times 10^{-8}\,\Omega\cdot\text{m}.
   $$

2. Substitute the values:  
   $$
   E = (1.72 \times 10^{-8}) \times (2.04 \times 10^{6}) \approx 3.51 \times 10^{-2}\,\text{V/m}.
   $$
   So,  
   $$
   E \approx 0.0351\,\text{V/m}.
   $$

---

**(c) Potential difference between two points 50 m apart**

1. The potential difference $$V$$ over a length $$L$$ is given by:  
   $$
   V = E L.
   $$
   
2. Here,  
   $$
   L = 50\,\text{m} \quad \text{and} \quad E \approx 0.0351\,\text{V/m}.
   $$

3. Therefore,  
   $$
   V = 0.0351\,\text{V/m} \times 50\,\text{m} \approx 1.755\,\text{V}.
   $$
   Hence,  
   $$
   V \approx 1.76\,\text{V}.
   $$

---

**(d) Resistance of a 50 m length of the wire**

1. The resistance $$R$$ is given by the formula:  
   $$
   R = \rho \frac{L}{A},
   $$
   where $$L = 50\,\text{m}$$ and the cross-sectional area  
   $$
   A \approx 8.174 \times 10^{-7}\,\text{m}^2.
   $$

2. Compute the numerator:  
   $$
   \rho L = 1.72 \times 10^{-8}\,\Omega\cdot\text{m} \times 50\,\text{m} = 8.6 \times 10^{-7}\,\Omega\cdot\text{m}^2.
   $$

3. Now, compute the resistance:  
   $$
   R = \frac{8.6 \times 10^{-7}}{8.174 \times 10^{-7}} \approx 1.052\,\Omega.
   $$
   Thus,  
   $$
   R \approx 1.05\,\Omega.
   $$
**(a) Current density:** The current density in the wire is approximately $$2.04 \times 10^{6}\,\text{A/m}^2$$. **(b) Electric field:** The electric field in the wire is approximately $$0.0351\,\text{V/m}$$. **(c) Potential difference:** The potential difference between two points 50 meters apart is approximately $$1.76\,\text{V}$$. **(d) Resistance:** The resistance of a 50-meter length of the wire is approximately $$1.05\,\Omega$$.

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