Answer:
Approximately 0.0852 Wb.
Explanation:
Find the radius of the circular coil from its circumference:
[tex]\displaystyle r = \frac{13.0}{30\times 2\pi}\; \rm m[/tex].
Find the area of this coil:
[tex]\displaystyle A = \pi r^{2} = {\frac{13.0}{30\times 2\pi}}^{2} = 0.00475647\; \rm m^{2}[/tex].
Find the magnetic flux linkage:
[tex]\displaystyle \phi= (B \times A) \cdot N = 0.00475647 \times 0.19 \times 30 \approx \rm 0.0852 \; Wb[/tex].
