A square loop of wire of side length L containing a load resistor R is oriented perpendicular to the xy-plane and rotates about the z-axis at an angular frequency omega in the presence of a magnetic field B=B_0 in the x-direction. If L=10 cm , B_0=2 T , and R= 100 Omega , what must omega be so that the average power dissipated, , is 1.0 W ?

The magnetic flux through the loop is the magnetic field times the component of the area vector that is parallel to field.
Phi_B=B_0*A cos(theta)=B_0*L^2 cos(omega t)
This generates an electromotive force in the wire.
epsilon=-d/dt Phi_B=-d/dt B_0*L^2 cos(omega t)=B_0 omega L^2 sin(omega t)
The power radiated by a resistor is:
P=V^2/R=P=epsilon^2/R=((B_0 omega L^2)^2 sin(omega t)^2)/R
The average power of a period is:
lt P gt =(B_0^2 omega^2 L^4)/(2R)
Solve for omega .
omega=sqrt((2RltPgt)/(B_0^2 L^4))=sqrt(2R ltPgt)/(B_0 L^2)
omega=sqrt(2*100 Omega*1 W)/(2 T* (0.1 m)^2)
omega=(sqrt(2)*10)/(0.02) s^-1
omega=sqrt(2)*500 s^-1 ~~707 s^-1