Correct Answer - A::B::C
(i) Induced emf
`E=-(d phi)/(dt)=-(d)/(dt)(BxxA)`
`=-(d)/(dt)[B(1/2 r^(2)theta)]=-1/2B(r^2)(d theta)/(dt)=-1/2B(r^2)(omega)`
`:. I=E/R=-1/2 (b(r^2)omega)/(R) implies|I|=1/2 (B(r^2)omega)/(R)`
(ii)The loop is entering in the magnetic field and hance magnetic lines of force passing through the loop is increasing in the downward direction. Therefore, current will flow in the loop in such a direction which will oppose the change. The current will flow in the anticlockwise direction.
(iii) Graph between induced enf and period of rotation: For first half rotation, (t=T//2), when the loop enters the fields, the current is in anticlockwise direction.
Magnitude of current remains constant at
`I=B(omega)(r^2)//(2R)`
`(B omega (r^2))/(2R)`
For next half rotation, when the loop comes out of the field, current of the same magnitude is set up clockwise.Anticlock current is supposed to be positive, The I-t graph is shown in the figure for two periods of rotation.