Calculating Average EMF in Two Coils on a Ferromagnetic Core

[James Holtum]

Hey guys been trying this question for a while now and i just can't seem to find any equations which link the two coils to find average EMF and also the final current I2. I am new to electrical so still learning any help is much appreciated, thanks!

Two coils, A and B, are wound on the same ferromagnetic core. There are 800 turns on A and 2800 turns on B. A current of 4 A through coil A produces a flux of 1500 μWb in the core. If this current is reversed in 30 ms, calculate the average EMFs induced in coils A and B.

Relevant equations
M=N2*(Flux2)/I1=N1*(Flux1)/I2

So for I've calculated M to be 1.05 mH and L to be 0.3 (not sure what units) but I don't think its right!

 

[Hesch]

Two coils, A and B, are wound on the same ferromagnetic core. There are 800 turns on A and 2800 turns on B. A current of 4 A through coil A produces a flux of 1500 μWb in the core. If this current is reversed in 30 ms, calculate the average EMFs induced in coils A and B.

Relevant equations
M=N2*(Flux2)/I1=N1*(Flux1)/I2

A and B is wound on the same core, so Flux1 = Flux2.

The equation to be used as for the Emf induced:

V = dΨ_n/dt , Ψ_n = Ψ * n = Flux * turns

 

[James Holtum]

So my value for N=2800+800?

Im given flux but that equation doesn't take into account the amperage and the time step?

Thanks

 

[Hesch]

There are 800 turns on A and 2800 turns on B. A current of 4 A through coil A produces a flux of 1500 μWb in the core. If this current is reversed in 30 ms, calculate the average EMFs induced in coils A and B.

So the current (4A) and thus the flux (1500 μWb) will change sign within 30 ms.

 

[rude man]

You don't need to compute inductances, including mutual inductance.
If 4A produces 1500 μWb in the core, and assume a linear change of current from +4A to -4A in 30 ms., what is dΦ/dt in the core? Then, what is the emf induced in each coil?
(The term "average emf" doesn't make sense to me. The "average" emf is as small as you want it to be by dividing by as long a time as you want. But if you assume a linear change in current you can use the resulting emf's as the answer).
What do you mean by I2? Current in coil B? That's obviously always zero. Final current in coil A? That's obvious.

 

[James Holtum]

You don't need to compute inductances, including mutual inductance.
If 4A produces 1500 μWb in the core, and assume a linear change of current from +4A to -4A in 30 ms., what is dΦ/dt in the core? Then, what is the emf induced in each coil?
(The term "average emf" doesn't make sense to me. The "average" emf is as small as you want it to be by dividing by as long a time as you want. But if you assume a linear change in current you can use the resulting emf's as the answer).
What do you mean by I2? Current in coil B? That's obviously always zero. Final current in coil A? That's obvious.

got 80 for A 280 for B

 

[rude man]

got 80 for A 280 for B

Very good, except 80 and 280 what?

 

[Hesch]

Very good, except 80 and 280 what?

@James Holtum: And remember that when calculating with AC-current in the coil, the induced voltage is the instant value, V(t), not the effective value, V_eff, or whatever.