Question

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

• A. 0.08 Wb
• B. 0.01 Wb
• C. 0.02 Wb
• D. 0.06 Wb

Answer 1

The Correct Option is C

$\overrightarrow{B}=\frac{1}{\pi}\big(\frac{Wb}{m^2}\big)$
Area of the disc normal to B is p$R^2$cos60$^{\circ}$.
Flux = B x Area normal
$\therefore \, Flux=\frac{1}{2}\times0.04=0.02 \, Wb$