Radiation from an Alternating Current Loop

This problem asks about the electric and magnetic fields produced by a circular wire loop carrying an alternating current in the far field, also known as the radiation zone. The given conditions are that the observation point \(\vec{r}\) is very far from the loop, specifically \(r \gg \frac{{\rm{c}}}{{\rm{\omega }}}\) >> R. The condition \(R \ll \frac{{\rm{c}}}{{\rm{\omega }}}\) means the loop radius is much smaller than the wavelength of the emitted radiation (\(\lambda = \frac{2\pi c}{\omega}\)). In this scenario, the loop behaves like an oscillating magnetic dipole.

Magnetic Dipole Moment

A circular current loop in the xy-plane carrying current \(I(t) = I_0\cos(\omega t)\) creates a magnetic dipole moment pointing along the z-axis (perpendicular to the loop). The magnitude of the magnetic dipole moment is \(m(t) = I(t) A\), where A is the area of the loop. For a circular loop of radius R, \(A = \pi R^2\). So, the magnetic dipole moment is \(\vec{m}(t) = m_0 \cos(\omega t) \hat{z}\), where \(m_0 = I_0 \pi R^2\).

Fields in the Radiation Zone

In the radiation zone (\(r \gg \frac{c}{\omega}\) or \(kr \gg 1\), where \(k = \omega/c\)), the electric and magnetic fields produced by an oscillating magnetic dipole \(\vec{m}(t) = m_0 \cos(\omega t) \hat{z}\) located at the origin are given in spherical coordinates \( (r, \theta, \phi) \) by:

Magnetic Field:

\(\vec{B}(\vec{r}, t) \approx \frac{\mu_0 m_0 k^2}{4\pi} \frac{\sin\theta}{r} \cos(\omega t - kr) \hat{\theta}\)

Electric Field:

\(\vec{E}(\vec{r}, t) \approx -\frac{\mu_0 m_0 k^2 c}{4\pi} \frac{\sin\theta}{r} \sin(\omega t - kr) \hat{\phi}\)

Here, \(\theta\) is the polar angle (angle with the z-axis, along which the dipole moment oscillates), and \(\phi\) is the azimuthal angle. \(k = \omega/c\) is the wave number.

Substituting \(m_0 = I_0 \pi R^2\) and \(k = \omega/c\), we get:

\(\vec{B}(\vec{r}, t) \approx \frac{\mu_0 (I_0 \pi R^2) (\omega/c)^2}{4\pi} \frac{\sin\theta}{r} \cos(\omega t - kr) \hat{\theta} = \frac{\mu_0 I_0 R^2 \omega^2}{4c^2} \frac{\sin\theta}{r} \cos(\omega t - kr) \hat{\theta}\)

\(\vec{E}(\vec{r}, t) \approx -\frac{\mu_0 (I_0 \pi R^2) (\omega/c)^2 c}{4\pi} \frac{\sin\theta}{r} \sin(\omega t - kr) \hat{\phi} = -\frac{\mu_0 I_0 R^2 \omega^2}{4c} \frac{\sin\theta}{r} \sin(\omega t - kr) \hat{\phi}\)

Analyzing the Options

Let's examine each statement based on these field expressions.

Option 1: The time-averaged \(\left| {\overrightarrow {{\rm{E}}\,} \,\left( {\overrightarrow {\rm{r}} ,{\rm{t}}} \right)} \right| \propto \frac{1}{{{r^2}}}\)

From the expression for \(\vec{E}(\vec{r}, t)\), the magnitude is:

\(|\vec{E}(\vec{r}, t)| = \left| -\frac{\mu_0 I_0 R^2 \omega^2}{4c} \frac{\sin\theta}{r} \sin(\omega t - kr) \right| = \left( \frac{\mu_0 I_0 R^2 \omega^2}{4c} \frac{|\sin\theta|}{r} \right) |\sin(\omega t - kr)|\)

The amplitude of the electric field is \(E_0 = \frac{\mu_0 I_0 R^2 \omega^2}{4c} \frac{|\sin\theta|}{r}\). The time-averaged value of \(|\vec{E}(\vec{r}, t)|\) is proportional to this amplitude. The amplitude \(E_0\) is proportional to \(1/r\), not \(1/r^2\). Therefore, Option 1 is incorrect.

Option 2: The time-averaged \(\left| {\overrightarrow {{\rm{E}}\,} \,\left( {\overrightarrow {\rm{r}} ,{\rm{t}}} \right)} \right| \propto \,{{\rm{\omega }}^2}\)

As shown above, the amplitude of the electric field is \(E_0 = \frac{\mu_0 I_0 R^2 \omega^2}{4c} \frac{|\sin\theta|}{r}\). This amplitude is directly proportional to \(\omega^2\). The time-averaged value of \(|\vec{E}(\vec{r}, t)|\) is proportional to \(E_0\). Therefore, the time-averaged \(|\vec{E}(\vec{r}, t)|\) is proportional to \(\omega^2\). This statement is correct.

Option 3: The time-averaged \({\overrightarrow {\rm{B}} \,\left( {\overrightarrow {\rm{r}} ,{\rm{t}}} \right)}\) as a function of the polar angle θ has a minimum at θ = π/2

The magnetic field is \(\vec{B}(\vec{r}, t) = \frac{\mu_0 I_0 R^2 \omega^2}{4c^2} \frac{\sin\theta}{r} \cos(\omega t - kr) \hat{\theta}\). The time-averaged value of the vector \(\vec{B}(\vec{r}, t)\) is zero, as it oscillates sinusoidally with time. If the option refers to the time-averaged magnitude squared (related to intensity), \(|\vec{B}(\vec{r}, t)|^2 \propto \left( \frac{\sin\theta}{r} \right)^2 \cos^2(\omega t - kr)\). The time average of \(\cos^2(\omega t - kr)\) is \(1/2\). So, the time-averaged \(|\vec{B}|^2\) is proportional to \(\sin^2\theta\). This term is maximum at \(\theta = \pi/2\) and zero at \(\theta = 0\) and \(\theta = \pi\). Thus, there is a maximum (of intensity/energy density) at \(\theta = \pi/2\), not a minimum. Option 3 is incorrect.

Option 4: \({\overrightarrow {\rm{B}} \,\left( {\overrightarrow {\rm{r}} ,{\rm{t}}} \right)}\) is along the azimuthal direction

From the expression for \(\vec{B}(\vec{r}, t)\), the magnetic field is along the \(\hat{\theta}\) direction (polar direction). The electric field \(\vec{E}(\vec{r}, t)\) is along the \(\hat{\phi}\) direction (azimuthal direction). Therefore, Option 4 is incorrect.

Based on the analysis of the fields in the radiation zone, only Option 2 is consistent with the derived expressions.