Average induced emf in a rotating coil after rotating by 180 degrees When a coil rotates in a magnetic field an emf is induced, and when the coil starts its motion from the position at which its plane is parallel to the field lines, and then it rotates by 180 degrees, the average induced emf in it is zero. I don't understand that point, and I would appreciate if someone explains it for me.
 A: For this question we must understand that the EMF produced is directly proportional to the rate of change of Magnetic flux linkage. 
As the field is parallel to the plane at first we can write that the magnetic flux density, B=k ( assuming be a constant).
Then the function of the flux through the coil will be Flux,F = BAsin(theta)=kAsin(theta) 
Note: We take sin as the function because we know t when field is parallel to coil, flux is zero so thus taking sin function which is zero when theta is zero.
Now EMF= -N(dF/dt)= -N(dF/dtheta )x(dtheta/dt) 
Now as you will rotate the coil at a constant angular velocity so dtheta/dt will be constant, assume it to be c( it is essentially angular velocity which remains constant as you are rotating at a constant speed here).
So now EMF=-Nc(d/dtheta(kAsin(theta))=-NckAcos(theta)
So the derivative of sin is cos so in interval 0 to 180, cos changes sign and has equal parts positive and negative so as the EMF is now a function of cos(theta) it's average value over interval from 0 to 180 degrees is 0

A: The EMF in the loop can be calculated using Faraday's law of induction, which states that the $V_{ems}$ is related to the time derivative of the magnetic flux through the loop:
$$V_{ems} = -\frac{d}{dt}\iint_{loop}\vec{B}.d\vec{\sigma}$$
In the case of a plane with surface area $A$ and a uniform angular velocity $\omega=\frac{2\pi}{T}$ in a homogeneous magnetic field $B$, the $V_{ems}$ will be:
$$V_{ems} = -\frac{d}{dt}\left(BA\sin{\omega t}\right) = -BA\omega\cos{\omega t}$$
When we want the average of some time-dependent quantity we calculate the integral of that quantity over the given time interval and divide by that time interval:
$$-BA\int_{0}^{\frac{T}{2}}\omega\cos(\omega t) dt = -BA\sin(\omega t)\Big|_0^\frac{T}{2}=0$$
Extra: intuitively this means that the induced current will flow in a certain way and direction during the first 90°; between $90°$and $180°$  the current will flow in the opposite direction, symmetric to the first $90°$, therefore the currents (or voltages if you like) will cancel each other out during every cycle of $180°$
