Question

Consider a tiny current loop driven by a sinusoidal alternating current. If the surface integral of its time-averaged Poynting vector is constant, then the magnitude of the time-averaged magnetic field intensity, at any arbitrary position, \(\mathbf{r}\), is proportional to

• A. \( \dfrac{1}{r^3} \)
• B. \( \dfrac{1}{r^2} \)
• C. \( \dfrac{1}{r} \)
• D. \( r \)

Hint:

In the far-field region of a current loop, the magnetic field intensity decreases with distance as \( \frac{1}{r} \), following the inverse proportionality rule.

Answer 1

The Correct Option is C

In electromagnetism, the Poynting vector represents the power flux (energy per unit time) in an electromagnetic field, and its time-averaged value is given by: \[ \langle \mathbf{S} \rangle = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}. \] Here, \( \mathbf{E} \) is the electric field, \( \mathbf{B} \) is the magnetic field, and \( \mu_0 \) is the permeability of free space. In the case of a tiny current loop driven by a sinusoidal alternating current, the time-averaged Poynting vector will be proportional to the rate of energy transfer. For a current loop, the time-averaged Poynting vector is constant, meaning that the energy carried by the electromagnetic waves emanating from the loop is uniform in all directions. This uniform distribution of energy means that the magnitude of the magnetic field intensity \( |\mathbf{B}| \) must decrease with distance from the loop. Using the relationship between the Poynting vector and the magnetic field intensity, it can be shown that the magnetic field intensity \( |\mathbf{B}| \) at a distance \(r\) from the current loop is proportional to \( \frac{1}{r} \). This is due to the inverse proportionality of the magnetic field intensity to the distance in cases of far-field propagation for sinusoidal oscillations. Thus, the magnitude of the time-averaged magnetic field intensity at any arbitrary position \(r\) is proportional to: \[ |\mathbf{B}| \propto \frac{1}{r}. \] Therefore, the correct answer is (C) \( \frac{1}{r} \).