A circular disc of radius 0.2 ; m is placed i | vedclass

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(C) The magnetic flux $\phi$ is given by the formula $\phi = \vec{B} \cdot \vec{A} = BA \cos 	heta$,where $	heta$ is the angle between the magnetic

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$0.02$

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(C) The magnetic flux $\phi$ is given by the formula $\phi = \vec{B} \cdot \vec{A} = BA \cos 	heta$,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 $R = 0.2 \; m$,so the area $A = \pi R^2 = \pi (0.2)^2 = 0.04\pi \; m^2$.The magnetic field $B = \frac{1}{\pi} \; Wb/m^2$.The axis of the disc makes an angle of $60^{\circ}$ with $\vec{B}$. Since the area vector $\vec{A}$ is along the axis of the disc,the angle $	heta$ between $\vec{B}$ and $\vec{A}$ is $60^{\circ}$.Therefore,$\phi = B A \cos 60^{\circ} = \left(\frac{1}{\pi}ight) 	imes (0.04\pi) 	imes \cos 60^{\circ}$.$\phi = 0.04 	imes \frac{1}{2} = 0.02 \; Wb$.

$A$ 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 $\vec{B}$. The magnetic flux linked with the disc is..... $Wb$.

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