Student Question

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, `<P>` , is `1.0 W` ?

Expert Answers

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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`

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