Answer
a. $1.0\times10^{-2}Wb$
b. $\theta = 48^{\circ}$
c. $6.7\times10^{-3}Wb$
Work Step by Step
a. When the plane of the loop is perpendicular to the magnetic field lines, the flux is given by the maximum value of equation 21–1. (There is a sign ambiguity.)
$$\Phi_B=BA=(0.50T)(\pi(0.08m)^2)=1.0\times10^{-2}Wb$$
b. When the plane of the loop makes a 42 degree angle with the field lines, the angle in equation 21-1 is given by $\theta = 90^{\circ}-42^{\circ}=48^{\circ}$.
c. Apply equation 21–1.
$$\Phi_B=BAcos\theta=(0.50T)(\pi(0.08m)^2) cos48^{\circ}=6.7\times10^{-3}Wb$$