A tightly-wound, long solenoid is initially kept with its axis...

The magnetic field produced by a large metal sheet carrying a surface current density $K$ is uniform and parallel to the sheet. By Ampere's law, its magnitude is given by:

$B_{\text{sheet}} = \dfrac{\mu_0 K}{2}$

Initially, the solenoid's axis is parallel to the sheet, and the net magnetic field at its centre is zero. This implies that the magnetic field produced by the solenoid exactly cancels the magnetic field of the sheet. Therefore, the magnetic field of the solenoid must be equal in magnitude and opposite in direction to the field of the sheet:

$B_{\text{sol}} = B_{\text{sheet}} = \dfrac{\mu_0 K}{2}$

When the solenoid is rotated such that its axis is perpendicular to the metal sheet, its magnetic field $B_{\text{sol}}$ becomes perpendicular to the sheet. However, the magnetic field of the sheet $B_{\text{sheet}}$ remains parallel to the sheet.

Since the two magnetic fields are now perpendicular to each other, the magnitude of the resultant magnetic field near the centre of the solenoid is the vector sum of the two fields:

$B_{\text{net}} = \sqrt{B_{\text{sol}}^2 + B_{\text{sheet}}^2}$

Substituting the values, we get:

$B_{\text{net}} = \sqrt{\left(\dfrac{\mu_0 K}{2}\right)^2 + \left(\dfrac{\mu_0 K}{2}\right)^2}$

$B_{\text{net}} = \sqrt{2 \left(\dfrac{\mu_0 K}{2}\right)^2} = \dfrac{\mu_0 K}{\sqrt{2}}$