First approach

The Ampere's force acting on a short piece of wire of length $\vec{\mathop{}\!\mathrm{d}l}$ is

\begin{equation}

\vec F_{amp} = I \vec{\mathop{}\!\mathrm{d}l} \times \vec B.

\end{equation}

The net force acting on this piece is

\begin{equation}

\quad\quad\quad\quad\quad\quad\quad\quad

\vec F = \vec T_1 + \vec T_2 + I\vec{\mathop{}\!\mathrm{d}l}\times \vec B = 0.

\quad\quad\quad\quad\quad\quad\quad\quad

(1)

\end{equation}

Here $\vec T_1$ and $\vec T_2$ are tension force of the wire acting on the two end points of the piece.

Projected on the $\vec{\mathop{}\!\mathrm{d}l}$, this equation reads:

\begin{equation}

T_1 = T_2.

\end{equation}

Therefore, the tension is constant along the wire.

Let $\vec n(l)$ be the tangent vector of the wire at the distance $l$ from the point of suspension. The equation (1) reads

\begin{equation}

\vec T_1 + \vec T_2 + I\vec{\mathop{}\!\mathrm{d}l}\times \vec B = \mathop{}\!\mathrm{d} l (T\frac{ \mathop{}\!\mathrm{d}{\vec n(l)}}{\mathop{}\!\mathrm{d}l} + I \vec n(l) \times \vec B) = 0,

\end{equation}

\begin{equation}

\frac{\mathop{}\!\mathrm{d}{\vec n(l)}}{\mathop{}\!\mathrm{d}l} = -\frac{I}{T} \vec n(l) \times \vec B.

\end{equation}

This equation implies that

\begin{equation}

\vec B \cdot \frac{\mathop{}\!\mathrm{d}{\vec n(l)}}{\mathop{}\!\mathrm{d}l} = 0 \implies \vec B \cdot n(l) = \text{const}.

\end{equation}

The tangent vector of the wire $\vec n$ is at constant angle to the magnetic field. In the horizontal plane the tangent vector is rotating at a constant speed. Thus, each side of the loop has the cylindrical helix shape, i. e. it is winding around some cylinder and making the constant angle $\alpha$ to the magnetic field. To find the radius of this cylinder, let's write the equation in the projection on the horizontal plane:

\begin{equation}

\frac{\mathop{}\!\mathrm{d}{\vec n_{xy}}}{\mathop{}\!\mathrm{d}l_{xy}}\sin(\alpha) = -\frac{I}{T}\vec n_{xy} \times \vec B.

\end{equation}

\begin{equation}

\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad

R = \frac{T}{IB}\sin(\alpha).

\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(2)

\end{equation}

At the point of the weight suspension we have the balance between three forces: two forces of tension(from each side of the loop) and the gravitational force of the weight, which is vertical. These three forces should be in the same vertical plane. Therefore, both sides of the loop are winding around the same cylinder. Each half of the loop should make a half-turn around this cylinder.

Thus, the length of the half of the loop is

\begin{equation}

L = \sqrt{H^2 + (\pi R)^2}.

\end{equation}

Therefore,

\begin{equation}

\quad\quad\quad\quad\quad\quad\quad

R = \frac{1}{\pi} \sqrt{L^2 - H^2} = \frac{1}{\pi} L \sin (\alpha).

\quad\quad\quad\quad\quad\quad\quad(3)

\end{equation}

The tension force of the wire can be found from equations (2) and (3)

\begin{equation}

T = \frac{IBR}{\sin(\alpha)} = \frac{I B L}{\pi}.

\end{equation}

The suspended weight is

\begin{equation}

P = 2T \cos(\alpha) = 2T \frac{H}{L}.

\end{equation}

\begin{equation}

P = \frac{2IBH}{\pi}.

\end{equation}

Second approach

Stable equilibrium corresponds to a minimum of potential energy. The potential energy is a sum of the gravitational energy of the weight and of the energy of the loop in the magnetic field. Let $S$ be an area of the loop projected on the horizontal plane.

\begin{equation}

E_{\text{p}} = -PH - ISB.

\end{equation}

$E_{\text{p}}$ depends on two variable parameters: $H$ and $S$. Note that we can change the form of the projection (not changing its length) of the loop onto horizontal plane without changing $H$. Thus, the minimum of the potential energy corresponds to the maximum area of the projection with fixed length — a circle.

Now, if the projection is fixed, the height $H$ is maximal if the wire makes a constant angle with the vertical axis. Thus, the wire has a shape of cylindrical helix — each side is winding a cylinder, while making constant angle $\alpha$ with vertical axis.

Let $R$ be the radius of the cylinder.

Then

\begin{equation}

S = \pi R^2 = \frac{1}{\pi} (L^2 - H^2).

\end{equation}

\begin{equation}

E_{\text{p}} = -PH - I B \frac{1}{\pi} (L^2 - H^2).

\end{equation}

The condition for the minimum of the potential energy is

\begin{equation}

\frac{\partial E}{\partial H} = 0.

\end{equation}

\begin{equation}

P = \frac{2IBH}{\pi}.

\end{equation}

The force balance equation at the point of weight suspension:

\begin{equation}

2 \frac{H}{L} T = P.

\end{equation}

\begin{equation}

T = \frac{LIB}{\pi}.

\end{equation}