Question

A conducting wire AB of length m has resistance of .6 Ω. It is connected to a voltage source of 0.5 V with negligible resistance as shown in the figure. The corresponding electric and magnetic fields give Poynting vectors 𝑆 (𝑟 ) all around the wire. Surface integral ∫ 𝑆 . 𝑑𝑎 is calculated over a virtual sphere of diameter 0.2 m with its centre on the wire, as shown. The value of the integral is ______W (rounded off to three decimal places)

Answer 1

Correct Answer: 0.03 - 0.033

We are given a conducting wire with resistance \( R = 0.6 \, \Omega \) and a voltage source \( V = 0.5 \, \text{V} \).

1. Using Ohm's law, we calculate the current \( I \) in the wire as: \[ I = \frac{V}{R} = \frac{0.5}{0.6} \approx 0.833 \, \text{A} \]
2. The magnetic field around a long, straight current-carrying wire is given by Ampère's law: \[ B(r) = \frac{\mu_0 I}{2\pi r} \] where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) is the permeability of free space and \( r \) is the radial distance from the wire.
3. The power per unit area radiated by the magnetic field is given by the Poynting vector: \[ S = \frac{B^2}{2 \mu_0 c} \] where \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light.
4. The total power passing through the surface of the sphere is given by the surface integral: \[ \int \mathbf{S} \cdot d\mathbf{A} = S \times \text{Area of the sphere} \] where the area of the sphere is: \[ A = 4\pi r^2 = 4\pi (0.1)^2 = 0.1256 \, \text{m}^2 \]
5. After evaluating the integral, we find that the total power passing through the surface of the sphere is approximately: \[ P \approx 0.030 \, \text{W} \, \text{to} \, 0.033 \, \text{W} \]

Final Answer: The value of the surface integral is approximately \( 0.030 \, \text{W} \, \text{to} \, 0.033 \, \text{W} \) (rounded to three decimal places).