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

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Aicharya · Accepted Answer

The magnetic flux $\phi$ passing surface of area $A$ placed in a u field $B$ is given by $\phi=B A \cos \theta$ where $\theta$ is the angle between

$\begin{aligned} & \text {Here, } \theta=60^{\circ}, B=\frac{1}{\pi} \mathrm{Wb} / \mathrm{m}^2, A=\pi(0.2)^2 \\ & \text {Therefore, } \phi=\frac{1}{\pi} \times \pi(0.2)^2 \times \cos 60^{\circ} \\ & =(0.2)^2 \times \frac{1}{2} \\ & =0.02 \mathrm{~Wb} \\ & \end{aligned}$

A circular disc of radius $0.2 \mathrm{~m}$ is placed in a uniform magnetic field of induction…

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