Answer:
1.13 Wb
Explanation:
First of all, we need to find the area enclosed by the coil.
The perimeter of the square is 2.0 m, so the length of each side is
[tex]L=\frac{2.0}{4}=0.5 m[/tex]
So the area enclosed by the coil is
[tex]A=L^2 = (0.5 m)^2=0.25 m^2[/tex]
Now we can calculate the magnetic flux through the square, which is given by
[tex]\Phi = B A cos \theta[/tex]
where
B = 9.0 T is the strength of the magnetic field
[tex]A=0.25 m^2[/tex] is the area of the coil
[tex]\theta[/tex] is the angle between the direction of the magnetic field and the normal to the coil; since the field is oriented 30.0° above the horizontal and the coil lies in the horizontal plane, the angle between the direction of the magnetic field and the normal to the coil is
[tex]\theta=90^{\circ}-30^{\circ}=60^{\circ}[/tex]
So the magnetic flux is
[tex]\Phi = (9.0)(0.25)(cos 60^{\circ})=1.13 Wb[/tex]
