Solution

The mutual inductance between the solenoid and the coil is obtained by calculating the magnetic flux of the solenoid through the coil.

The magnetic field inside the long solenoid is uniform and axial: $B = \mu_0 n I$, where I is the current in the solenoid and n is the number of turns per unit length.

The bent coil is formed by two half-turns of radius r, each with area$S = \pi r^2/2$.

Their planes make angles $\alpha$and $\frac{\pi}{2}-\alpha$ with the solenoid axis, so their normals make angles $\frac{\pi}{2}-\alpha$and $\alpha$, respectively, with the direction of the field. The flux through each half-turn is $B\,S\cos\theta$, where $\theta$ is the angle between the normal and the field. Thus, the total flux is:

$\Phi = B\frac{\pi r^2}{2}\left[\cos\!\left(\frac{\pi}{2}-\alpha\right) + \cos\alpha\right] = \mu_0 n I\frac{\pi r^2}{2}(\sin\alpha + \cos\alpha)$

The mutual inductance is $M = \Phi/I$ So

$\boxed{M = \frac{1}{2}\,\mu_0 n \pi r^2 (\sin\alpha + \cos\alpha)}$