The induced emf in a straight wire
General form of the equation: \$ V_{Ind} = N B l v \ sin\ \theta\$

Where:


\$ V_{Ind}\$ = Induced Voltage in V.
\$N\$ = Number of turns - in your case 1.
\$B\$ = Flux Density in T (Teslas = Wb/m^2).
\$v\$ = velocity in m/s.
\$l\$ = length in m.


Let \$ V_{MAX} = N B l v\$.

Motion can be split into two components, \$x = V_{MAX}\ cos\ \theta\$ and \$y = V_{MAX}\ sin\ \theta\$.

The x-component (\$cos\ \theta\$) would be in parallel with the magnetic field.  No flux lines are crossed, no induced EMF.  It can be ignored.

The y-component (\$sin\ \theta\$) is perpendicular to magnetic field.  It is the only component which crosses the lines of magnetic flux, so it is the component which induces the voltage by Faraday's Law.  Maximum induced voltage would occur at \$90\unicode{xb0}\$, and min at \$0\unicode{xb0}\$.

This demonstrates the how motion of a wire in a magnetic field induces a voltage.  The basis for generators.  Too get any meaningful voltage you will have to have many turns.

Now, look at your drawing and concentrate on the motion.  Horizontal movement would produce no voltage (parallel to magnetic field).  Vertical maximum.  This should help you clarify which angle \$\theta\$ or \$\theta _L\$.