Answer:
83.3 Wb
Explanation:
The magnetic flux linkage through the coil is given by:
[tex]N\phi = BAN sin \theta[/tex]
where
B is the magnetic field strength
A is the cross sectional area
N is the number of turns in the coil
[tex]\theta[/tex] is the angle between the direction of the field and the normal to the coil
In this problem:
B = 1.74 T
A = 0.133 m^2
N = 360
[tex]\theta=90^{\circ}[/tex]
Therefore, the magnetic flux linkage is
[tex]N\phi = (1.74 T)(0.133 m^2)(360) sin 90^{\circ}=83.3 Wb[/tex]
