Time Varying Fields and Maxwells Equations Questions and Answers

Answer: A

A helical winding closed round the conductor links the flux of its field, and the induced voltage is proportional to the rate of change of the enclosed current. Integrating it recovers the current waveform, and because there is no iron the device never saturates.

Answer: B

A coil carrying alternating current induces currents within the workpiece, which dissipate in its resistance. The frequency is chosen so that the skin depth suits the depth of heating wanted, high frequency for surface hardening and low for through heating.

Answer: C

The changing flux induces an emf in the turn and its low resistance lets a large current flow, which both heats it and opposes the flux. An accidental shorted turn in a transformer core clamp is a well-known cause of local overheating.

Answer: D

Each turn separately encircles the flux, so the flux linkage is N times the flux and the emf is N times that of a single turn. It is the same reason the inductance of a coil goes as the square of the turns, one factor for producing the flux and one for linking it.

Answer: A

The double line integral makes the reciprocity of mutual inductance obvious, since exchanging the two circuits merely exchanges the order of integration. It is rarely the easiest way to compute a mutual inductance but it is the most illuminating.

Answer: B

With no resistance any emf would drive an unbounded current, so the total flux linkage cannot change; the induced current adjusts to hold it constant. This flux conservation is the basis of superconducting magnets and of the persistent currents observed in them.

Answer: C

Both contain a time derivative of a flux density, but only Ampere's has a source current, because no magnetic current of monopoles exists. That single asymmetry runs through the whole of the theory and is the deepest structural difference between electricity and magnetism.

Answer: D

The equation says that current out of a closed surface equals the rate of decrease of the charge within, which is conservation of charge in local form. Energy has its own conservation statement in Poynting's theorem, which is the subject of a separate topic.

Answer: A

The electric field at the surface is \(\frac{1}{2}\mu_{0}NR\omega\alpha\sin\omega t\) in magnitude.

Inside the solenoid \(B = \mu_{0}NI = \mu_{0}N\alpha\cos\omega t\). Faraday’s law around a circle of radius \(R\) gives

\[ \begin{aligned} E(2\pi R) &= -\frac{d}{dt}(\pi R^{2}B) \\ &= \pi R^{2}\mu_{0}N\alpha\omega\sin\omega t \\ E &= \tfrac{1}{2}\mu_{0}NR\omega\alpha\sin\omega t \end{aligned} \]

The options with \(\cos\omega t\) forget that the induced field depends on the rate of change of the current, which brings in \(\sin\omega t\); they also drop the amplitude \(\alpha\). \(\mu_{0}NR\omega\alpha\sin\omega t\) misses the factor \(\tfrac{1}{2}\) that comes from dividing the area \(\pi R^{2}\) by the circumference \(2\pi R\).

Answer: D

The radiated power is proportional to \(q^{2}\omega^{4}A^{2}\).

Larmor's formula gives the power radiated by an accelerating charge as \(P = \frac{q^{2}a^{2}}{6\pi\epsilon_{0}c^{3}}\), proportional to the square of both the charge and the acceleration. For simple harmonic motion \(x = A\cos\omega t\), the acceleration amplitude is \(\omega^{2}A\), so

\[ P \propto q^{2}(\omega^{2}A)^{2} = q^{2}\omega^{4}A^{2} \]

The options with a single power of \(q\) forget that power goes as the field squared, and those with \(\omega^{2}\) use the velocity amplitude \(\omega A\) instead of the acceleration.