But from first relation, $V_s \propto \frac {1} {I_s}$
Actually, no. We have:
$N_s = N_p \dfrac{I_p}{I_s}$
So, $V_s \propto \frac {1} {I_s}$ is not true since $I_p$ is not constant.
The secondary voltage and current are proportional by Ohm's Law:
(1) $V_s = R_L I_s$
The power is conserved:
(2) $V_p I_p = V_s I_s = R_L I^2_s$
The secondary current is related to the primary current:
(3) $I_s = \dfrac{N_p}{N_s}I_p$
Thus:
$V_p I_p = R_L (N_p/N_s)^2 I^2_p$
$\rightarrow \dfrac{V_p}{I_p} = (N_p/N_s)^2R_L$
I will answer this as if this is an Electrical Engineering course or site. We electrical engineers use "$j$" for the imaginary unit: $j^2 = -1$. We use the small-case "$i$" to represent current in the time domain. We use small case for signals in the time domain (such as $v(t)$ and $i(t)$) and we use large case of the same letter (and subscript) to indicate the same signals in the frequency domain ($V(j\omega)$, $I(j\omega)$) or equivalently Laplace Transforms of the same signals ($V(s)$, $I(s)$).
And this is the convention for defining the voltages and currents in a two-port mutual inductance (otherwise known as a transformer). The dot convention $\bullet$ simply means that the two inductors, as seen looking at the wires marked with the dot are wound around the common core in the same sense. Let's say they both have the same "right-hand rule".
Given these definitions and conventions, the time-domain volt-ampere equations for the mutual inductance is:
$$\begin{align}
v_1(t) &= L_1 \frac{d}{dt}i_1(t) + M \frac{d}{dt}i_2(t) \\
v_2(t) &= M \frac{d}{dt}i_1(t) + L_2 \frac{d}{dt}i_2(t) \\
\end{align}$$
The mutual inductance, $M$, is the same value in both equations and is related to the two individual inductances, $L_1$ and $L_2$, as
$$ 0 \le M = k \sqrt{L_1 L_2} \le \sqrt{L_1 L_2} $$
$k$ is the "coefficient of coupling", which is always between 0 and 1.
$$ 0 \le k \le 1 $$
Also, given the dimensions of a simple cylindrical wound ideal inductor, the inductance is:
$$ L = \mu N^2 \pi r^2 \ell = \mu N^2 (\text{Vol}) $$
$\mu$ is the permeability of the core, $N$ is the number of turns of winding the wire (BTW, the wire must be at least thinly insulated, often they are "enameled") and $r$ and $\ell$ are the cross section radius and length of the cylinder. The cross-sectional area is $\pi r^2$ and the volume is $ \text{Vol} = \pi r^2 \ell $. The salient fact to remember is that the inductance is proportional to the square of the number of turns.
The Laplace Transforms of the two equations are:
$$\begin{align}
V_1(s) &= s \, L_1 I_1(s) + s \, M I_2(s) \\
V_2(s) &= s \, M I_1(s) + s \, L_2 I_2(s) \\
\end{align}$$
or the frequency-domain representation is (substituting $s \leftarrow j\omega$):
$$\begin{align}
V_1(j \omega) &= j \omega \, L_1 \, I_1(j \omega) + j \omega \, M \, I_2(j \omega) \\
V_2(j \omega) &= j \omega \, M \, I_1(j \omega) + j \omega \, L_2 \, I_2(j \omega) \\
\end{align}$$
I am not finding the perfect image on the web of an ideal transformer with a load and keeping with the current and voltage conventions I have above. So I am using this picture:
Now keeping in the Laplace domain, we can add two equations:
$$\begin{align}
V_g(s) &= R_1 I_1(s) + V_1(s) \\
V_2(s) &= -I_2(s) R_2 \\
\end{align}$$
The reason why we show $-I_2$ is because current is defined as positive flowing into $L_2$ and that is opposite of (or negative of) the current flowing into $R_2$. And we don't really give a rat's ass about $V_g$.
So solving these three equations
$$\begin{align}
V_1(s) &= s \, L_1 I_1(s) + s \, M I_2(s) \\
V_2(s) &= s \, M I_1(s) + s \, L_2 I_2(s) \\
V_2(s) &= -I_2(s) R_2 \\
\end{align}$$
for $V_1$ in terms of $V_2$ we get
$$\begin{align}
I_2(s) &= \frac{-1}{R_2} V_2(s) \\
\\
I_1(s) &= \frac{V_2(s) - s L_2 I_2(s)}{sM} \\
&= \frac{V_2(s) + \frac{s L_2}{R_2} V_2(s)}{sM} \\
\\
V_1(s) &= s \, L_1 \frac{V_2(s) + \frac{s L_2}{R_2} V_2(s)}{sM} + s \, M \frac{-1}{R_2} V_2(s) \\
&= \left(\frac{1}{k}\sqrt{\frac{L_1}{L_2}} + s \left(\frac{1}{k} - k \right)\frac{\sqrt{L_1 L_2}}{R_2} \right) V_2(s)
\end{align}$$
Now, in the ideal case, when the coefficient of coupling, $k$, approaches 1, then the second term on the right goes to 0 and we have
$$ V_1(s) = \sqrt{\frac{L_1}{L_2}} V_2(s) $$
and $\sqrt{\frac{L_1}{L_2}}$ is real and positive and no $s$ factor in it. If the two coils are interwound over the same core and share the same dimensions (same $r$ and same $\ell$) then the square root of the ratio of inductances is the same as the turns ratio of the primary-to-secondary windings which is $\frac{N_1}{N_2}$. So
$$ V_1(s) = \frac{N_1}{N_2} V_2(s) $$
After inverse Laplace transforming, we have
$$ v_1(t) = \frac{N_1}{N_2} v_2(t) $$
Best Answer
A significant source of power loss in a transformer is the induced eddy currents in the core. Just as the varying magnetic field induces current in the secondary coil, it can also induce currents in the core itself. These currents do nothing but dissipate energy, and so are to be avoided.
To reduce eddy currents, you either build your transformer out of a non-conductive magnetic material (e.g. ferrite), or you split the core's conductive material into many plates separated by insulating layers. The insulating layers block the large-scale eddy currents while passing the magnetic field, thus reducing the power loss.