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A circular disc of radius 0.2 metre is placed in a uniform magnetic field of induction $\dfrac{1}{\pi}\left(\dfrac{Wb}{m^2}\right)$ in such a way that its axis makes an angle of $60^\circ$ with $\vec{B}$. The magnetic flux linked with the disc is -
A
0.08 Wb
B
0.01 Wb
C
0.02 Wb
D
0.06 Wb
Detailed Solution
Area of the disc: $A = \pi r^2 = \pi (0.2)^2 = 0.04\pi$ m$^2$
The axis of the disc is along its area vector, so the angle between $\vec{A}$ and $\vec{B}$ is $\theta = 60^\circ$.
Magnetic flux: $\phi = BA\cos\theta$
$\phi = \dfrac{1}{\pi} \times 0.04\pi \times \cos 60^\circ$
$\phi = 0.04 \times \dfrac{1}{2}$
$\phi = 0.02$ Wb
The axis of the disc is along its area vector, so the angle between $\vec{A}$ and $\vec{B}$ is $\theta = 60^\circ$.
Magnetic flux: $\phi = BA\cos\theta$
$\phi = \dfrac{1}{\pi} \times 0.04\pi \times \cos 60^\circ$
$\phi = 0.04 \times \dfrac{1}{2}$
$\phi = 0.02$ Wb