A circular disc of the radius $0.2 ext{ m}$ is placed in a uniform magnetic field of induction $\frac{1}{\pi

Question

A circular disc of the radius $0.2 	ext{ m}$ is placed in a uniform magnetic field of induction $\frac{1}{\pi} 	ext{ Wb m}^{-2}$ in such a way that its axis makes an angle of $60^\circ$ with $\vec{B}$. The magnetic flux linked to the disc will be:

Accepted Answer

The magnetic flux ($\Phi_B$) through a surface is defined as the scalar product of the magnetic field ($\vec{B}$) and the area vector ($\vec{A}$): 
$$ \Phi_B = \vec{B} \cdot \vec{A} = BA \cos 	heta $$

1.  **Identify parameters:**
    *   Magnetic Field ($B$) = $\frac{1}{\pi} 	ext{ Wb m}^{-2}$
    *   Radius ($r$) = $0.2 	ext{ m}$
    *   Area ($A$) = $\pi r^2 = \pi (0.2)^2 = 0.04\pi 	ext{ m}^2$
    *   Angle ($	heta$): The area vector $\vec{A}$ is directed along the normal to the surface, which coincides with the axis of the disc. The problem states the axis makes an angle of $60^\circ$ with $\vec{B}$. Therefore, $	heta = 60^\circ$ .

2.  **Calculation:**
    Substituting the values into the flux formula:
    $$ \Phi_B = \left( \frac{1}{\pi} ight) (0.04\pi) \cos(60^\circ) $$
    $$ \Phi_B = 0.04 	imes 0.5 $$
    $$ \Phi_B = 0.02 	ext{ Wb} $$

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