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.
In general, a inductive 'loop' for picking up a magnetic field (as shown connected to the galvanometer in you diagram) is 'symmetric' with respect to rotation about the axis parallel to the magnetic field from the coil connected to the battery. If the 'loop' is square (as shown), the symmetry is perfect for 90 degree rotations. There may be some small variation in the magnetic flux enclosed by a 'square' coil rotated at 45 degrees for example. A circular loop rotated about it's centre would be a better chioce.