Time Varying Fields and Maxwells Equations Questions and Answers

Answer: B

The Lorentz gauge condition is \(\vec{\nabla}\cdot\vec{A}\) \(+ \mu_{0}\varepsilon_{0}\,\partial V/\partial t = 0\).

The Lorentz (Lorenz) gauge chooses the potentials so that the divergence of \(\vec{A}\) cancels the time derivative of V divided by \(c^{2}\). With \(c^{2} = 1/\mu_{0}\varepsilon_{0}\):

\[ \begin{aligned} &\vec{\nabla}\cdot\vec{A} \\ &+ \frac{1}{c^{2}}\frac{\partial V}{\partial t} = 0 \end{aligned} \]

In this gauge both V and \(\vec{A}\) satisfy inhomogeneous wave equations, and the condition is Lorentz invariant. Writing \(1/\mu_{0}\varepsilon_{0}\) instead multiplies by \(c^{2}\) where it should divide, the minus sign gives the wrong relative sign, and the second time derivative belongs to the wave equation, not the gauge condition.

Answer: B

The radiated power is minimum along the y-axis.

A current loop in the xz-plane has its magnetic dipole moment along the normal to the loop, the \( y \)-axis. Like an electric dipole, an oscillating magnetic dipole radiates with an intensity proportional to \( \sin^{2}\theta \), where \( \theta \) is measured from the dipole axis. So no power goes out along the axis itself.

The \( x \)-axis, the \( z \)-axis and the direction \( x\hat{x} + z\hat{z} \) all lie in the plane of the loop, at \( \theta = 90^{\circ} \) from the moment, where the radiation is maximum.

Answer: B

The equation represents the continuity equation.

It expresses conservation of charge. Integrated over a volume and using the divergence theorem, it says that the current flowing out through the surface equals the rate at which the charge inside decreases:

\[ \oint \vec{j}\cdot d\vec{a} = -\frac{d}{dt}\int \rho\,dV \]

The wave equation is second order in time and space, \(\nabla^{2}\psi = \frac{1}{v^{2}}\frac{\partial^{2}\psi}{\partial t^{2}}\). Poisson's equation, \(\nabla^{2}\varphi = -\rho/\varepsilon_{0}\), relates potential to charge density, and Laplace's equation is its charge-free form. Neither involves current density.

Answer: A

The current in the coil increases.

By Faraday's law the induced emf is proportional to the rate of change of flux linkage, \(e = -N\,d\phi/dt\). If the magnet moves through the coil at a steady speed, the flux changes at a steady rate and the emf and current are roughly constant. If the magnet is accelerating, its speed keeps rising, so the flux changes faster and faster and the induced emf, and hence the current in a closed coil, keeps increasing.

The current would decrease if the magnet were slowing down, would be constant at steady speed, and reverses only when the magnet's motion reverses or it passes out of the far end; acceleration in one direction does none of these.

Answer: C

The ratio of conduction to displacement current density is the loss tangent.

In a medium with conductivity \(\sigma\) and permittivity \(\varepsilon\), a time-harmonic field drives a conduction current density \(\sigma E\) and a displacement current density \(j\omega\varepsilon E\). Their magnitudes give

\[ \begin{aligned} \tan\theta &= \frac{|J_{c}|}{|J_{d}|} \\ &= \frac{\sigma}{\omega\varepsilon} \end{aligned} \]

which is the loss tangent. It classifies media: much greater than 1 for a good conductor, much less than 1 for a good dielectric.

The attenuation constant \(\alpha\) and propagation constant \(\gamma\) describe how a wave decays and travels, and the dielectric constant is \(\varepsilon_{r}\) alone; they depend on the loss tangent but are not that ratio.

Answer: A

The law that does not form a Maxwell equation is Planck's law.

Maxwell's four equations are Gauss's law for electricity, Gauss's law for magnetism (no magnetic monopoles), Faraday's law of induction, and Ampere's circuital law with Maxwell's displacement current added. Together they describe all classical electromagnetic fields and predict electromagnetic waves.

Planck's law describes the spectrum of black-body radiation and introduced the quantisation of energy, \(E = h\nu\). It belongs to thermal radiation and quantum physics and is not one of Maxwell's equations.

Answer: C

The statement is true when the Lorentz gauge condition \(\partial A^{\nu}/\partial x^{\nu} = 0\) is satisfied.

In terms of the potentials, the inhomogeneous Maxwell equations read

\[ \begin{aligned} \Box A^{\mu} - \partial^{\mu}(\partial_{\nu}A^{\nu}) \\ = -\mu_{0}J^{\mu} \end{aligned} \]

where \(\Box = \nabla^{2} - \frac{1}{c^{2}}\frac{\partial^{2}}{\partial t^{2}}\). Only when \(\partial_{\nu}A^{\nu} = 0\) does the extra term vanish, leaving the four decoupled wave equations of the question. (The homogeneous equations are satisfied automatically once the fields are written as derivatives of \(A^{\mu}\).)

So it is not true in any gauge. The Coulomb condition \(\vec{\nabla}\cdot\vec{A} = 0\) decouples the equations differently, leaving an instantaneous Poisson equation for the scalar potential. The first option is wrong because the homogeneous equations need no separate wave equation.

Answer: A

The wave equation is \(\nabla^2 E = \dfrac{1}{c^2}\dfrac{\partial^2 E}{\partial t^2}\), which is item 4, so a matches 4. The cross product of E and H is the Poynting vector, the power flow per unit area, so b matches 3. Gauss's law in differential form is \(\nabla \cdot D = \rho\), item 2, so c matches 2. Ampere's law with Maxwell's correction is \(\nabla \times H = J + \partial D/\partial t\), item 1, so d matches 1. The code is 4 3 2 1. The code 3 4 2 1 swaps the first two, and the codes 1 2 3 4 and 4 2 3 1 mismatch Gauss's and Ampere's laws. The answer is 4 3 2 1.