How is $B_p$ directly proportional to $I_p$ in a transformer?

Question

The textbook I am studying from states the following

The voltage $V_P$ in the primary coil is directly proportional to the current $I_p$ in the coil $$V_p\propto I_p$$ The magnetic flux density $B_p$ is also directly proportional to $I_p$ $$B_p\propto I_p$$

However I do not understand why exactly or which equation links the magnetic flux density to the current in that way, all I could come up with based on reasoning is that,

Since the induced E.M.F is given by$$E=BLV\sin(\theta)$$ Hence $$E\propto B$$ And so since $V_p\propto I_p$ and $E\propto B$ hence $$B \propto I_p$$ However I know that this is a hackjob at most , how is the proper way to go about thinking of this relationship in terms of proportionalities?

Answer 1

I suppose it is AC current. In this case the variation of the magnetic field induces a tension: $$V \propto \frac{\partial B}{\partial t}$$

But the magnetic field is a result of the current: $B \propto I$

So, $$V \propto \frac{\partial I}{\partial t}$$