Question

A resistor consists of a solid cylinder of radius \(r\) and length \(L\). The resistor has resistance \(R\) and is carrying current \(i\). Use the Poynting vector to calculate the power radiated out of the surface of the resistor

Short answer

Express your answer in terms of r, L, R, i, and the permittivity of free space (μ₀). Answer: The power radiated out of the surface of a cylindrical resistor is given by \(P = \frac{2\pi rIR}{\mu_0}\).

Step-by-step solution

Find the electric field in the resistor

Since the resistor is carrying a current, there will be an electric field present within the resistor. The electric field in the resistor can be calculated by Ohm's Law: \(V = IR\). We can write this as \(\vec{E} = \frac{IR}{L}\), where \(\vec{E}\) is the electric field in the resistor directed along its length.

Find the power density using the Poynting vector

To find the power radiated out of the surface of the resistor, we will now use the Poynting vector, which represents the power per unit area. In this case, since the magnetic field is negligible, the Poynting vector is almost entirely due to the electric field. The Poynting vector is given by \(\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B}\). Since the magnetic field is negligible, we can approximate the Poynting vector as \(\vec{S} \approx \frac{1}{\mu_0} \vec{E}\).

Calculate the power radiated out of the surface

Now that we have the Poynting vector, we can calculate the power radiated out of the surface of the resistor. Recall that the Poynting vector represents the power per unit area, so we can find the total power by integrating the Poynting vector over the surface of the resistor. The surface area of the resistor's curved part is given by \(A_s = 2\pi rL\). Therefore, the total power radiated out of the surface of the resistor is given by: \(P = \int_A \vec{S} \cdot \mathrm{d}\vec{A}\), where A is the surface area of the resistor. Since \(\vec{S} \approx \frac{1}{\mu_0} \vec{E}\), we can plug that into the equation: \(P = \int_A \frac{1}{\mu_0} \vec{E} \cdot \mathrm{d}\vec{A}\). Substituting the expression for \(\vec{E}\) found in step 1 and the surface area of the resistor: \(P = \frac{1}{\mu_0} \int_{A_s} \frac{IR}{L} \cdot \mathrm{d}A\). Integrating over the surface area: \(P = \frac{IR}{\mu_0 L} \cdot 2\pi rL = \frac{2\pi rIR}{\mu_0}\). Thus, the power radiated out of the surface of the resistor is given by \(P = \frac{2\pi rIR}{\mu_0}\).