The magnetic flux ($\Phi_B$) through a surface is defined as the scalar product of the magnetic field ($\vec{B}$) and the area vector ($\vec{A}$): $\Phi_B = \vec{B} \cdot \vec{A} = BA \cos \theta$

1. Identify parameters: Magnetic Field ($B$) = $\frac{1}{\pi} \text{ Wb m}^{-2}$ Radius ($r$) = $0.2 \text{ m}$ Area ($A$) = $\pi r^2 = \pi (0.2)^2 = 0.04\pi \text{ m}^2$ Angle ($\theta$): The area vector $\vec{A}$ is directed along the normal to the surface, which coincides with the axis of the disc. The problem states the axis makes an angle of $60^\circ$ with $\vec{B}$. Therefore, $\theta = 60^\circ$ .

2. Calculation: Substituting the values into the flux formula: $\Phi_B = \left( \frac{1}{\pi} \right) (0.04\pi) \cos(60^\circ)$ $\Phi_B = 0.04 \times 0.5$ $\Phi_B = 0.02 \text{ Wb}$