How Is EMF Generated in a Coil Flipped in a Magnetic Field?

[Moh'd Ali MQN]

A 25-turn circular coil of wire has diameter 1.00 m. It is placed with its axis along the direction of the Earth’s magnetic field of 50.0 /T, and then in 0.200 s it is flipped 180°. An average emf of what magnitude is generated in the coil?

This is a question in my book and the answer was
http://store1.up-00.com/Jun11/M8h57751.p…
Why didnt we use the formula where we derive the angle as : E=-N.d(BAcos(wt))/dt
My solution was as follows:
E = -NBAw.sin(wt), where w = 2.pi/t
E = -NBAw.sin(180) = 0 and this was logical to me because the final emf in the coil is zero and we get an average of zero. Unless there is another meaning of average emf...

help me out please

 

[Gordianus]

Earth's magnetic field is well below 50 T. As a matter of fact I don't know whether such a strong field has been achieved yet. Anyway, let´s say you have an inmensely powerful magnet that provides a uniform B of 50 T. When the coil's axis points along B, the magnetic flux across the coil is computed as B*A* N, where A is the area of the coil and N the number of turns. When the coil is flipped 180 degrees the flux has the same magnitude but opposite sign because the axis is pointing the other way. Thus the flux has changed 2*B*A*N. Know, you divide the flux change into the time interval and you get the average induced electromagnetic force.