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(B) The magnetic flux $\phi$ linked with a surface in a uniform magnetic field is given by $\phi = B A \cos 	heta$, where $	heta$ is the angle betwe

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$0.02$

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(B) The magnetic flux $\phi$ linked with a surface in a uniform magnetic field is given by $\phi = B A \cos 	heta$, where $	heta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$ (which is normal to the surface).
Given, radius $r = 20 	ext{ cm} = 0.2 	ext{ m}$.
Area $A = \pi r^2 = \frac{22}{7} 	imes (0.2)^2 = \frac{22}{7} 	imes 0.04 	ext{ m}^2$.
Magnetic field $B = \frac{7}{22} 	ext{ Wb/m}^2$.
The angle between the axis of the disc and $\vec{B}$ is $60^{\circ}$. Since the area vector $\vec{A}$ is along the axis of the disc, the angle $	heta$ between $\vec{A}$ and $\vec{B}$ is $60^{\circ}$.
Therefore, $\phi = B A \cos 60^{\circ} = \left( \frac{7}{22} ight) 	imes \left( \frac{22}{7} 	imes 0.04 ight) 	imes 0.5$.
$\phi = 0.04 	imes 0.5 = 0.02 	ext{ Wb}$.

$A$ circular disc of radius $20 \text{ cm}$ is placed in a uniform magnetic field of induction $\frac{7}{22} \text{ Wb/m}^2$ in such a way that its axis makes an angle of $60^{\circ}$ with $\vec{B}$. The magnetic flux linked with the disc is $(\cos 60^{\circ} = 0.5)$ (in $\text{ Wb}$)

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