Answer:
0.04 Wb
Explanation:
The flux through the coil at a generic time t is given by
[tex]\Phi = BA cos \theta[/tex]
where
B = 454 mT = 0.454 T is the magnetic field intensity
A is the area of the coil
[tex]\theta[/tex] is the angle between the direction of the field B and the normal to the surface of the coil
The side length of the coil is
L = 83.0 cm = 0.83 m
so its area is
[tex]A=L^2=(0.83 m)^2=0.69 m^2[/tex]
At the initial instant, the angle is
[tex]\theta=30^{\circ}[/tex]
so the magnetic flux through the coil is
[tex]\Phi_i = BAcos \theta_i = (0.454 T)(0.69 m^2)(cos 30^{\circ})=0.27 Wb[/tex]
At the final instant, the angle is
[tex]\theta=0^{\circ}[/tex]
so the magnetic flux through the coil is
[tex]\Phi_f = BAcos \theta_i = (0.454 T)(0.69 m^2)(cos 0^{\circ})=0.31 Wb[/tex]
So, the change in magnetic flux is
[tex]\Delta \Phi = \Phi_f - \Phi_i = 0.31 Wb-0.27 Wb=0.04 Wb[/tex]
