Problem 24.11

24.11. a)
Emf is given by
$\epsilon = - \frac{d\Phi_{B}}{dt}$
$\epsilon = - \frac{d}{dt} (B A \cos \theta)$
Where the angle $\theta$ is the angle that the normal of the loop (area vector) makes with the magnetic field.
$\epsilon = - A \cos \theta \frac{dB}{dt}$
The magnetic field sent out by the TV station takes the form
$B = B_{m} \cos \omega t$
Where $\omega$ is given by
$\omega = 2\pi f$
So, we have
$\epsilon = -A \cos \theta \frac{d}{dt} B_{m} \cos (2\pi f t)$
$\epsilon = - A B_{m} \cos \theta \frac{d}{dt} \cos (2\pi f t)$
But we know that
$\frac{d}{dt} \cos (2\pi f t) = - (2\pi f) \sin (2\pi f t)$
So, we have
$\epsilon (t) = A B_{m} (2\pi f) \cos \theta \sin (2\pi f t)$
Finally, we know that the area is given by
$A = \pi r^{2}$
So, that gives us
$\epsilon (t) = 2\pi^{2} r^{2}f B_{m} \cos \theta \sin (2\pi f t)$

b)
If the electric field is vertical, then the magnetic field is horizontal. The plane of the loop should then be vertical so that the magnetic field passes through the area of the loop. Furthermore, the loop should be turned so that the normal to the loop (area vector) is parallel or antiparallel to the direction of the transmitting antenna.