(a) When bar magnet is moved towards left, the anticlockwise current will be in loop (1) and (2) {facing bar magnet} following Lenz’s law. Thus, the direction of induced current in loop (1) will be from A to B and that in (2) will be from D to C. 

(b) When bar magnet is moved towards right, the clockwise current will be in loop (1) and (2) {facing bar magnet} following Lenz’s law. Thus, the direction of induced current in the loop (1) will be from B to A and C to D.

Given, a coil of N turns having an area A, rotated with angular speed '$\mathrm{\omega }$'  in a uniform magnetic field 'B' connected to a resistor 'R' .
 
The flux linking to the coil,  $\boxed{\mathrm{\varphi } = \mathrm{NBA} \sin \left(\mathrm{\omega t}\right)}$ 
Therefore,
Induced EMF, $$\begin{gathered} \mathrm{E} = \frac{\mathrm{d}\left(\mathrm{NBA} \cos \mathrm{\omega t}\right)}{\mathrm{dt}} \\ = \mathrm{NBA} \mathrm{\omega } \sin \mathrm{\omega t} \end{gathered}$$  

When, sin$\mathrm{\omega }$t = $\pm$1, the induced emf value is maximum. 

(i) The maximum EMF, ${\mathrm{E}}_0 = \mathrm{NBA} \mathrm{\omega }$ 

(ii) The power dissipated in the coil is given by,  
                 $$\begin{gathered} \mathrm{P}= \frac{{{\mathrm{E}}^2}_{\mathrm{rms}}}{\mathrm{R}} \\ = \frac{{\mathrm{E}}_0^2}{2\mathrm{R}} \\ = \frac{(\mathrm{NAB} \mathrm{\omega }{)}^2}{2\mathrm{R}} \end{gathered}$$