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:
...
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Already a member? Log in here.
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`
