Energy In A Magnetic Cycle
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Energy In A Magnetic Cycle
During each magnetic cycle, the energy expended in the specimen is proportional to the area of the closed loop.
Consider a ring of specimen of mean circumference l metres, cross-sectional area a metres2 and having N turns of an insulated wire. Let the current flowing through the wire be of I amperes.
Magnetising force, H = NI
l
or I = H l
N
Let the flux density at this instant be B.
Total flux through the ring, ∅ = B X a
If the current is increased to increase the magnetizing force H and induction density B.
Induced e.m.f. in the coil.
e = - Number of turns on coil x rate of change of flux
= - N d∅ = - Nd(Ba) = - Na dB
dt dt dt
According to Lenz’s law this induced e.m.f. will oppose the flow of current, therefore in order to maintain the current , therefore in order to maintain the current I in the solenoid, the source of supply must have an equal and opposite e.m.f.
Hence applied e.m.f. e = - Na dB
dt
Energy consumed in short time dt = e x I x dt = Na dB X I dt
dt
= Na dB X Hl x dt = a l H d B since I = Hl
dt N N
Total energy consumed =
Now al is the volume of ring and HdB is the area of elementary strip of B-H curve shown in HdB is the total area enclosed by the hysteresis loop.
Energy consumed/cycle = Volume of the ring x area of the loop
Energy consumed per cycle per cubic metre of volme = area of the hysteresis lop
This energy expended in taking a specimen through a magnetic cycle is wasted and since it appears as heat, so it is termed as hysteresis loss.
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Consider a ring of specimen of mean circumference l metres, cross-sectional area a metres2 and having N turns of an insulated wire. Let the current flowing through the wire be of I amperes.
Magnetising force, H = NI
l
or I = H l
N
Let the flux density at this instant be B.
Total flux through the ring, ∅ = B X a
If the current is increased to increase the magnetizing force H and induction density B.
Induced e.m.f. in the coil.
= - N d∅ = - Nd(Ba) = - Na dB
dt dt dt
According to Lenz’s law this induced e.m.f. will oppose the flow of current, therefore in order to maintain the current , therefore in order to maintain the current I in the solenoid, the source of supply must have an equal and opposite e.m.f.
Hence applied e.m.f. e = - Na dB
dt
Energy consumed in short time dt = e x I x dt = Na dB X I dt
dt
= Na dB X Hl x dt = a l H d B since I = Hl
dt N N
Total energy consumed =
Now al is the volume of ring and HdB is the area of elementary strip of B-H curve shown in HdB is the total area enclosed by the hysteresis loop.
Energy consumed/cycle = Volume of the ring x area of the loop
Energy consumed per cycle per cubic metre of volme = area of the hysteresis lop
This energy expended in taking a specimen through a magnetic cycle is wasted and since it appears as heat, so it is termed as hysteresis loss.
For more help in Energy In A Magnetic Cycle click the button below to submit your homework assignment