Energy in transformer air gap
It is the other way around, the magnetic current is constant (Flux and B = Flux/Area) in the core. And the magnetic voltage (magnetic field strength H) will drop across the core length (\$I * N = H_c*l_c+H_g*l_g\$). 

And we will have the larges "magnetic voltage drop" across the larges magnetic resistance (reluctance). The ferrite core will have lower magnetic reluctance compared to the air gap magnetic reluctance. 

$$ R = \frac{l}{\mu_o\mu_r A}$$ 

So if the core length is \$l_e = 37mm\$, the Area of a core is \$A_e = 20mm^2\$ 

And the air gap is length \$l_g = 0.1mm\$   

We have the air gap magnetic reluctance will be equal to about if we ignore the Flux fringing. 

$$R_g \approx \frac{0.1mm}{4\pi 10^{-7} 20mm^2} \approx 398000 \:H^{-1}$$ 

The ferrite core magnetic reluctance will be equal to if the relative permeability of a ferrite core is \$\mu_r = 2000\$  

$$R_c \approx \frac{37mm}{4\pi 10^{-7} *2000* 20mm^2} \approx 73609\:H^{-1}$$ 

Therefore the magnetic voltage drop across the air gap will be 398000/73609 = 5.4 times large in the air gap compared to the core magnetic voltage drop.   

Or we can use the fact that the flux (the magnetic current) is the same in the caore is is inr the air gap we will have:

$$ B = \mu_o \mu_r H_c = \mu_o H_g  $$

Thus:

$$H_g = H_c \mu_r$$

This means that the magnetic field strength in the air gap will be larger than the \$Hc\$ in the core.

In our example, we have \$\mu_r = 2000\$ and the air gap length is \$370\$ times smaller than the core length. The magnetic field strength in the air gap will be \$ \frac{2000}{370} \approx 5.4 \$ large than the H in the core.