A) [tex]1.22\cdot 10^{-5} Wb[/tex]
The total magnetic flux through the coil before it is rotated is given by:
[tex]\Phi_i = NBA cos \theta[/tex]
where
N = 180 is the number of turns in the coil
[tex]B=5.00\cdot 10^{-5}T[/tex] is the magnetic field intensity
[tex]A=13 cm^2 = 13\cdot 10^{-4} m^2[/tex] is the area enclosed by the coil
[tex]\theta=0^{\circ}[/tex] is the initial angle between the direction of the normal to the surface of the coil and the magnetic field
Substituting, we find
[tex]\Phi_i = (180)(5.00\cdot 10^{-5} T)(13.5\cdot 10^{-4} m^2)(cos 0^{\circ})=1.22\cdot 10^{-5} Wb[/tex]
B) 0
The total magnetic flux through the coil after it is rotated is still given by:
[tex]\Phi_i = NBA cos \theta[/tex]
where this time,
[tex]\theta=90^{\circ}[/tex] is the final angle between the direction of the normal to the surface of the coil and the magnetic field
since [tex]cos \theta=cos 90^{\circ} =0[/tex], we now have that the total flux through the coil is zero.
C) [tex]2.44\cdot 10^{-4} V[/tex]
The magnitude of the average emf induced in the coil is given by
[tex]\epsilon=\frac{\Delta \Phi}{\Delta t}[/tex]
where
[tex]\Delta \Phi = \Phi_i =1.22\cdot 10^{-5}Wb[/tex] is the change of total flux
[tex]\Delta t=5.00\cdot 10^{-2} s[/tex] is the time taken
Susbtituting, we find
[tex]\epsilon=\frac{1.22\cdot 10^{-5} Wb}{5.00\cdot 10^{-2}s}=2.44\cdot 10^{-4} V[/tex]
