A circular disc of area $(4\hat{i} + 5\hat{j}) \times 10^{-3}m^2$ is placed in a uniform magnetic field of intensity $(0.2 \hat{i} + 0.3\hat{j})$Tesla. The flux crossing the disc will be-
Answer & explanation
Correct answer: option 4
$\phi = \vec{B}.\vec{A} = (0.2 \hat{i} + 0.3\hat{j}). (4\hat{i} + 5\hat{j}) \times 10^{-3} = 0.0008 + 0.0015 = 23\times 10^{-4}$