Question

A voltage, \(V\), is applied across a cylindrical conductor of radius \(r\) length \(L,\) and resistance \(R .\) As a result, a current, \(i,\) is flowing through the conductor, which gives rise to a magnetic field, \(B\). The conductor is placed along the \(y\) -axis, and the current is flowing in the positive \(y\) -direction. Assume that the electric field is uniform throughout the conductor. a) Find the magnitude and the direction of the Poynting vector at the surface of the conductor of the static electric and magnetic fields. b) Show that \(\int \vec{S} \cdot d \vec{A}=i^{2} R\)

Short answer

Question: Show that the \(\int \vec{S} \cdot d \vec{A} = i^{2} R\) for a cylindrical conductor with given radius, length, and resistance when a voltage is applied along the conductor causing a current to flow and generating a magnetic field. Answer: To show that \(\int \vec{S} \cdot d \vec{A} = i^{2} R\), we first calculate the electric field (E) inside the conductor using Ohm's law, which is given by E = V / L, where V is the voltage and L is the length of the conductor. Then, we determine the magnetic field (B) using Ampere's law, B = (μ₀*i) / (2*pi*r), where μ₀ is the permeability constant and i is the current flowing through the conductor. Next, we find the Poynting vector (S) using the formula S = (1/μ₀) * E x B and substitute the expressions for E and B we derived earlier, resulting in a radial component (outward) from the surface of the conductor. Finally, we evaluate the integral of the Poynting vector over the surface of the conductor and show that \(\int \vec{S} \cdot d \vec{A} = i^{2} R\).

Step-by-step solution

Electric Field Calculation

In order to find the Poynting vector, we first need to determine the electric field inside the conductor. Since the electric field is uniform throughout the conductor with resistance R and the voltage V is applied across the length L, we can use Ohm's law to calculate the electric field. V = E * L Where E is the electric field, rearrange the equation for electric field: E = V / L

Magnetic Field Calculation

Next, we need to calculate the magnetic field (B) generated due to the current flowing through the conductor. We will make use of Ampere's law to find the magnetic field: B*2*pi*r = μ₀*i Where μ₀ is the permeability constant and i is the current flowing through the conductor. Rearrange the equation for the magnetic field: B = (μ₀*i) / (2*pi*r)

Poynting Vector Calculation

Now that we have both the electric field and magnetic field, we can determine the Poynting vector. The Poynting vector (S) is given by: S = (1/μ₀) * E x B As the conductor is placed along the y-axis, the electric field (E) is directed in the positive y-direction and the magnetic field (B) is in the azimuthal direction (tangential to the surface of the conductor). The cross product of E and B results in the Poynting vector having a radial component (pointing outwards from the surface of the conductor). Therefore, the magnitude of the Poynting vector is: S = (1/μ₀) * E * B Now substitute the expressions for E and B we derived earlier: S = (1/μ₀) * (V / L) * ((μ₀*i) / (2*pi*r)) Since 1/μ₀ = μ₀/(4*pi), we have: S = (μ₀*i*V) / (4*pi*r*L) The direction of the Poynting vector is radial (outward) from the surface of the conductor.

Evaluating the Integral

To show that \(\int \vec{S} \cdot d \vec{A} = i^{2} R\), we need to evaluate the integral of the Poynting vector over the surface of the conductor. Since both the Poynting vector and the surface vector are radial, their dot product will only consider their magnitudes: \(\int \vec{S} \cdot d \vec{A} = \int S * dA\) The surface area (dA) of the conductor can be written as: dA = L * 2*pi*r * dr Integrating the Poynting vector over the surface of the conductor: \(\int \vec{S} \cdot d \vec{A} = \int_{0}^{r} (μ₀*i*V) / (4*pi*r*L) * (L * 2*pi*r) dr\) Simplifying the expression, we get: \(\int \vec{S} \cdot d \vec{A} = (μ₀*i*V) * \int_{0}^{r} dr\) Now integrate the right-hand side: \(\int \vec{S} \cdot d \vec{A} = (μ₀*i*V) * [r]_{0}^{r} = μ₀*i*V*r\) Since we know that V = i * R, we can substitute this into the equation: \(\int \vec{S} \cdot d \vec{A} = μ₀*i^{2}*R*r\) From the above expression for B: B = (μ₀*i) / (2*pi*r), we can find the value of the constant μ₀: μ₀ = (2*pi*r*B) / i Substitute this value into the equation above: \(\int \vec{S} \cdot d \vec{A} = ((2*pi*r*B) / i) * i^{2}*R*r = i^{2} R\) And this completes the proof.