Answer:
B_o = 1.013μT
Explanation:
To find B_o you take into account the formula for the emf:
[tex]\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}[/tex]
where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.
By applying the derivative you obtain:
[tex]\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)[/tex]
when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:
[tex]\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T[/tex]
hence, B_o = 1.013μT