Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Moving Charges and Magnetism

Question:

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-

Options:

23 Weber

$23\times 10^{-2} Weber$

$23\times 10^{-3} Weber$

$23\times 10^{-4} Weber$

Correct Answer:

$23\times 10^{-4} Weber$

Explanation:

$\phi = \vec{B}.\vec{A} = (0.2 \hat{i} + 0.3\hat{j}). (4\hat{i} + 5\hat{j}) \times 10^{-3} = 0.0008 + 0.0015 = 23\times 10^{-4}$