Answer:
54 mWb
Explanation:
The magnetic flux linkage through the coil is given by:
[tex]\Phi = BA sin \theta[/tex]
where
B is the magnetic field strength
A is the cross sectional area
[tex]\theta[/tex] is the angle between the direction of the field and the normal to the coil
In this problem:
B = 0.92 T
A = 0.074 m^2
[tex]\theta=52^{\circ}[/tex]
Therefore, the magnetic flux linkage is
[tex]\Phi = (0.92 T)(0.074 m^2) sin 52^{\circ}=0.054 Wb=54 mWb[/tex]
