Given:
The radius of the circular loop is,
[tex]\begin{gathered} r=18.761\text{ cm} \\ =18.761\times10^{-2}\text{ m} \end{gathered}[/tex]The magnetic field is,
[tex]B=4.751\text{ T}[/tex]The normal to the plane of the loop is parallel to the field.
To find:
The magnetic flux
Explanation:
The magnetic flux is given as,
[tex]\Phi=BAcos\theta[/tex]Here, A is the area of the loop and
[tex]\theta[/tex]is the angle between the area vector of the loop and the magnetic field.
As we know, the area vector is always perpendicular to the plane of the loop, so the angle in this case is,
[tex]\theta=0\degree\text{ \lparen parallel\rparen}[/tex]The area of the loop is,
[tex]\begin{gathered} A=\pi r^2 \\ =\frac{22}{7}\times18.761\times18.761\times10^{-4} \\ =0.1106\text{ m}^2 \end{gathered}[/tex]The magnetic flux is,
[tex]\begin{gathered} \Phi=4.751\times0.1106\times cos0\degree \\ =0.525\text{ Wb} \end{gathered}[/tex]Hence, the magnetic flux is 0.525 Wb.
