Question

A circular disc of radius $0.2\, m$ is placed in a uniform magnetic field of induction $\frac{1}{\pi}\left(\frac{ Wb }{ m ^{2}}\right)$ in such a way that its axis makes an angle of $60^{\circ}$ with B. The magnetic flux linked with the disc is:

• A. $0.02\, Wb$
• B. $0.06\, Wb$
• C. $0.04\, Wb$
• D. $0.01\, Wb$

Answer 1

Answer: (A)

$\phi=B A \cos \theta$

$ \therefore \phi =\frac{1}{\pi} \times \pi(0.2)^{2} \times \cos 60^{\circ} $
$=(0.2)^{2} \times \frac{1}{2}=0.02\, Wb $