Magnetostatics and Amperes Law Questions and Answers

Answer: D

It may equally be written as \( \tfrac{1}{2}\mathbf{B} \cdot \mathbf{H} \), and integrating it over all space gives the same total as the inductance expression. At one tesla in iron the density is of order four hundred kilojoules per cubic metre, far more than any electrostatic field achieves.

Answer: A

Both losses rise steeply with frequency, so an iron core that is efficient at fifty hertz becomes a resistor at a megahertz. Ferrites, which are ceramic and of high resistivity, fill the gap where a core is needed at radio frequency.

Answer: B

An Amperian circle drawn inside the hollow encloses no current, so the line integral of H is zero and the symmetry forces the field to vanish. This is the magnetic counterpart of the field-free interior of a charged conducting shell.

Answer: C

The field intensity is set by the ampere-turns per metre and is unchanged, but the flux density is multiplied by the relative permeability. Because that permeability falls as the iron saturates, the relation between current and flux is no longer a straight line.

Answer: D

A shell of mumetal offers a path of very low reluctance, so the flux prefers it to the enclosed space and the interior field is much reduced. Against a steady field conductivity is useless, since no eddy currents flow, which is the opposite of the electrostatic case.

Answer: A

Fifty microteslas, or half a gauss, is the usual figure, and it is worth carrying because it sets the scale against which stray fields and screening requirements are judged. A conductor carrying twenty amperes matches it at about a hundred millimetres.

Answer: B

The force is developed on the current-carrying conductor and is passed to the core through the insulation and wedges that hold it in the slot, which is why slot wedging is a mechanical design matter. A stationary conductor in a field experiences the force just as a moving one does.

Answer: C

Each coil links its own flux and that of the other, so the mutual term appears twice. Reversing one coil makes the fluxes oppose and the mutual term is subtracted twice instead, which is the standard method of measuring mutual inductance from two series readings.

Answer: D

One reading is the sum of the self inductances plus twice the mutual and the other the same minus twice the mutual, so the difference is four times the mutual inductance. Dividing the measured difference by four is therefore the whole of the measurement.