Askiitians Tutor Team
Last Activity: 4 Months ago
To tackle this problem, we need to analyze the given conductor's properties and how they relate to the electric field and resistance. The conductor has a circular cross-section, a specific length, and a potential difference applied across its ends. Let's break this down step by step.
Understanding the Parameters
We have a solid conductor with the following specifications:
- Radius (r) = 5 mm = 0.005 m
- Length (L) = 20 m
- Potential difference (V) = 0.1 V
- Magnetic field intensity (H) = 10^5 A/m²
Finding the Electric Field (E)
The relationship between the electric field (E) and the magnetic field intensity (H) in a conductor can be described using Ohm's law in its differential form:
J = σE, where J is the current density, σ is the conductivity, and E is the electric field.
Since we have a uniform magnetic field, we can relate the current density to the magnetic field intensity:
J = H * σ
From the above equations, we can express E in terms of H and σ:
E = J / σ = (H * σ) / σ = H
Thus, the electric field E is equal to the magnetic field intensity H in this case.
Calculating the Resistance (R)
The resistance of a conductor can be calculated using the formula:
R = ρ * (L / A)
Where:
- ρ = resistivity (inverse of conductivity, ρ = 1/σ)
- L = length of the conductor
- A = cross-sectional area of the conductor
The cross-sectional area A of a circular conductor is given by:
A = πr² = π(0.005 m)² = π(25 x 10^-6 m²) ≈ 7.85 x 10^-5 m²
Finding Conductivity (σ)
To find the conductivity σ, we need to relate it to the electric field and the potential difference:
Using Ohm's law, V = E * L, we can express E as:
E = V / L = 0.1 V / 20 m = 0.005 V/m
Now, we can find the conductivity σ using the relationship:
σ = J / E = (H * σ) / E
From our previous calculations, we know H = 10^5 A/m² and E = 0.005 V/m. Rearranging gives us:
σ = H / E = 10^5 A/m² / 0.005 V/m = 2 x 10^7 S/m
Calculating Resistivity (ρ)
Now that we have σ, we can find the resistivity ρ:
ρ = 1 / σ = 1 / (2 x 10^7 S/m) = 5 x 10^-8 Ω·m
Final Calculation of Resistance (R)
Substituting the values into the resistance formula:
R = ρ * (L / A) = (5 x 10^-8 Ω·m) * (20 m / 7.85 x 10^-5 m²)
Calculating this gives:
R ≈ (5 x 10^-8) * (254.65) ≈ 1.27 x 10^-5 Ω
Summary of Results
In summary, the conductivity σ as a function of radius is constant in this case, approximately 2 x 10^7 S/m, and the resistance between the two ends of the conductor is approximately 1.27 x 10^-5 Ω. This analysis illustrates how the properties of the conductor interact with the applied potential difference and the resulting electric field.
