Physics going on in a transformer

So I read your responses which left me thinking further about the original question.


Would I be correct in stating the following and if so please feel free to expand in greater detail, or correct, though having read Mikes last post I think you may have already beaten me to the punch.

With no load on the secondary there is a 'magnetising current in the primary the size of which depends upon the design and build quality of the transformer. This primary current creates the magnetic field around the core, alternating back and forth. 


The strength of the magnetic field is given by the equation, B = u*n*l (permeability * number of turns * current) 


The energy in the magnetic field is 



Therefore the amount of energy in the magnetic field is dependent upon the size of the primary current. Power is the rate of energy conversion which for the primary will be electric to magnetic energy at voltage * current. There will be a loss of energy due to Lenz's law across the primary coil.


The magnetic field induces an emf on the secondary coil. This when loaded will draw a current. This current too will as a result of Lenz's law result in a further partial loss of energy from the magnetic field.


This induced magnetic field as a result of the current on the secondary actually counteracts the total size of the magnetic field from the primary, and therefore the energy in it.


If you could take an instant snap shot in time and at the same time not cause a change in the primary current this reduction in the size of the magnetic field from the primary means the amount of energy in the magnetic field available has to decrease too. The rate of this reduction has to be matched by the rate of energy withdrawal from the magnetic field at the secondary side which explains why with voltage fixed the current has to increase in a step down transformer. 


In reality until the transformer is fully loaded to its maximum Kva rating the current in the primary will increase putting energy back into the magnetic field to maintain the energy levels which with its maximum available size per given current has to have a matching energy withdrawal at the secondary side. 


​​​​​​​Neil

So I read your responses which left me thinking further about the original question.


Would I be correct in stating the following and if so please feel free to expand in greater detail, or correct, though having read Mikes last post I think you may have already beaten me to the punch.

With no load on the secondary there is a 'magnetising current in the primary the size of which depends upon the design and build quality of the transformer. This primary current creates the magnetic field around the core, alternating back and forth. 


The strength of the magnetic field is given by the equation, B = u*n*l (permeability * number of turns * current) 


The energy in the magnetic field is 



Therefore the amount of energy in the magnetic field is dependent upon the size of the primary current. Power is the rate of energy conversion which for the primary will be electric to magnetic energy at voltage * current. There will be a loss of energy due to Lenz's law across the primary coil.


The magnetic field induces an emf on the secondary coil. This when loaded will draw a current. This current too will as a result of Lenz's law result in a further partial loss of energy from the magnetic field.


This induced magnetic field as a result of the current on the secondary actually counteracts the total size of the magnetic field from the primary, and therefore the energy in it.


If you could take an instant snap shot in time and at the same time not cause a change in the primary current this reduction in the size of the magnetic field from the primary means the amount of energy in the magnetic field available has to decrease too. The rate of this reduction has to be matched by the rate of energy withdrawal from the magnetic field at the secondary side which explains why with voltage fixed the current has to increase in a step down transformer. 


In reality until the transformer is fully loaded to its maximum Kva rating the current in the primary will increase putting energy back into the magnetic field to maintain the energy levels which with its maximum available size per given current has to have a matching energy withdrawal at the secondary side. 


​​​​​​​Neil