Q. A circular disc of radius $0.2\ m$ is placed in a uniform magnetic field of induction $\frac{1}{\pi}(Wb/m^{2})$ in such a way that its axis makes an angle of $60{^\circ}$ with vector $\overrightarrow{B}$. The magnetic flux linked with the disc is
A) 0.08 $Wb$
B) 0.01 $Wb$
C) 0.02 $Wb$
D) 0.06 $Wb$
Explanation
(c) $r = 0.2\ m,\ B = \frac{1}{\pi},\ Q = 60{^\circ},\ a = ?$
$\phi = BA\cos\theta = B\left( \pi r^{2} \right)\cos\theta$
$= \frac{1}{\pi}(\pi \times 0.2 \times 0.2)\cos{60{^\circ}} = 0.02Wb$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.