MHTCET 20 Apr 2026 Shift 2 Paper

Concept:Magnetic flux is the product of magnetic field, area, and the cosine of the angle between the field and the normal to the surface.Explanation:For a circular disc, the axis is perpendicular to its plane, so the axis itself is the normal to the disc.The angle given ($60^\circ$) is between the axis and $\vec{B}$, so this is directly the angle $\theta$ in the flux formula.Convert the radius to metres: $r = 20\ \text{cm} = 0.20\ \text{m}$.Area of the disc: $A = \pi r^2 = \pi (0.20)^2 = 0.04\pi\ \text{m}^2$.Given $B = \frac{7}{22}\ \text{Wb/m}^2$ and $\cos 60^\circ = 0.5$.Magnetic flux: $\phi = B A \cos \theta$.Substitute the values: $\phi = \frac{7}{22} \times 0.04\pi \times 0.5$.Using $\pi = \frac{22}{7}$, the $\frac{7}{22}$ and $\pi$ cancel, giving $\phi = 0.04 \times 0.5 = 0.02\ \text{Wb}$.Thus, the magnetic flux linked with the disc is $0.02\ \text{Wb}$.Answer:$0.02\ \text{Wb}$ — Option B.