Question: ![heres a diagram of a solenoid and another coil connected to the galvanometer](https://i.stack.imgur.com/lL5Q7.jpg)

If the coil on the left is rotated about the magnetic field lines, will there be an induced current? 
I know that the flux concerns only the component of the magnetic field perpendicular to the plane of the coil, so even if it is rotated, I assume the perpendicular component will remain the same. Just want to confirm this. Thanks

Answer: 
The flux is defined as integal {if you don't know calculus, take this as a sum} of $\int\_\Gamma \vec B \dot \, \mathrm d \vec A$ where dA points out the surface $\Gamma$.

[![](https://i.stack.imgur.com/M7WXF.gif)](https://i.stack.imgur.com/M7WXF.gif)  

(source: [gsu.edu](http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/imgmag/flux2.gif))

if you had a circular shape of coil, the flux would not change, and hence, no Electric field or EMF would be induced, but here, as the field of solenoid may not be perfectly symmetrical outside it, (spreads out evenly in each direction) . hence, there is bound to be some emf, not too much, theoretically for an ideal solenoid, there would be no emf induced too.
Depends on the angular velocity of its rotation. Electromagnetic interactions propagates at light speed. This means, if the coil is rotating in a frequency low enough, and if the induction circuit is close enough, the magnetic fields will arrive in the other circuit, almost "statically". Therefore, as good approximation in this case, you can consider the magnetic field arriving in the induction circuit, arrived "instantaneously" from the coil.

If coil has length $L$, rotating at angular speed $\omega$, and the induction circuit is at a distance $s$ from the coil, then our approximations: The extreme part of the coil must have speed lower than speed of light: $L\omega\ll c$. And the time for the magnetic field arrive the circuit must be close to zero: $\Delta t = s/c\ll 1$. If this is true, you can calculate the emf with induction law approximating that the magnetic field rotates nicely with the coil:
$$
\mbox{emf} = -\frac{d\Phi}{dt}
$$