ES2C6 Sample Paper –Solutions
N.B. As a design paper there will be some questions that open-ended, the idea being that I create an
opportunity for you to show your understanding without expecting you to report a single, particular
fact or number. As such some of the questions have multiple ‘right’ answers, E.g., Q3c where you
are asked “with the aid of a suitably detailed diagram…”. The solutions offered here then are an
indication of the depth I’d expect to win marks in various areas. In places I have referred to the
lecture slides as these show the underlying principle which is probably more useful than a set form
of words in most cases. Also I think it is important you review the material provided carefully as
there are a lot of good examples in the various slides including many worked examples.
Hope this helps.
RL
Q3
a) Lecture 24
Magneto-motive force : in ampere-turns (AT). Analogous to EMF in an electrical circuit. The source
of magnetizing energy – the potential for the device to produce flux. May be determined via :
ℱ= ℛ = NI = HL
Flux : in weber. ‘Magnetic current’. Magnetic energy ‘flow’ through the iron circuit. Analogous to
current in an electrical circuit.
Reluctance: in A/Wb or per H. The reluctance to the ‘flow’ of flux for a given magnetizing MMF.
Analogous to resistance in an electrical circuit.
b) Lecture 30 p33 on is helpful.
The trick here is to recognize that the arms form the sides that make the gap, so there is a coil of
2000 turns wrapped around a fat core, then two skinnier arms coming out of the ends that don’t
quite meet and form a 0.5 mm gap.
Start with the flux : B = phi / area → phi = B x area = 0.9 T x 60 mm2 = 54 x 10-6 Wb.
Flux is conserved; there is no fringing / no losses so the flux through all circuit elements is the same.
The field in the gap is given as 0.9 T. The gap has the same XSA as the arms so the field in the arms is
equal to this (as B=phi/area and phi and area are the same). Bgap = Barm1 = Barm2 = 0.9 T.
The field in the core is different, being less as the area is greater :
Bcore = phi / area = 54 x 10-6 Wb / 80 mm2 = 0.675 T.
Using the look-up table given find Harms and Hcore.
By interpolation : Harm = 650 AT/m; Hcore = 400 AT/m
The field in gap is given, and as its air you know μ0 and so can work out Hgap.
Hgap = Bgap / μ0 = 716 200 AT/m
N.B. As a design paper there will be some questions that open-ended, the idea being that I create an
opportunity for you to show your understanding without expecting you to report a single, particular
fact or number. As such some of the questions have multiple ‘right’ answers, E.g., Q3c where you
are asked “with the aid of a suitably detailed diagram…”. The solutions offered here then are an
indication of the depth I’d expect to win marks in various areas. In places I have referred to the
lecture slides as these show the underlying principle which is probably more useful than a set form
of words in most cases. Also I think it is important you review the material provided carefully as
there are a lot of good examples in the various slides including many worked examples.
Hope this helps.
RL
Q3
a) Lecture 24
Magneto-motive force : in ampere-turns (AT). Analogous to EMF in an electrical circuit. The source
of magnetizing energy – the potential for the device to produce flux. May be determined via :
ℱ= ℛ = NI = HL
Flux : in weber. ‘Magnetic current’. Magnetic energy ‘flow’ through the iron circuit. Analogous to
current in an electrical circuit.
Reluctance: in A/Wb or per H. The reluctance to the ‘flow’ of flux for a given magnetizing MMF.
Analogous to resistance in an electrical circuit.
b) Lecture 30 p33 on is helpful.
The trick here is to recognize that the arms form the sides that make the gap, so there is a coil of
2000 turns wrapped around a fat core, then two skinnier arms coming out of the ends that don’t
quite meet and form a 0.5 mm gap.
Start with the flux : B = phi / area → phi = B x area = 0.9 T x 60 mm2 = 54 x 10-6 Wb.
Flux is conserved; there is no fringing / no losses so the flux through all circuit elements is the same.
The field in the gap is given as 0.9 T. The gap has the same XSA as the arms so the field in the arms is
equal to this (as B=phi/area and phi and area are the same). Bgap = Barm1 = Barm2 = 0.9 T.
The field in the core is different, being less as the area is greater :
Bcore = phi / area = 54 x 10-6 Wb / 80 mm2 = 0.675 T.
Using the look-up table given find Harms and Hcore.
By interpolation : Harm = 650 AT/m; Hcore = 400 AT/m
The field in gap is given, and as its air you know μ0 and so can work out Hgap.
Hgap = Bgap / μ0 = 716 200 AT/m
