Answer:
0.3405V
Explanation:
#Given a magnetic field of [tex]1.9T[/tex], diameter= 18.5cm(r=9.25cm or 0.0925m), we find the magnetic flux of the loop as:
[tex]\phi=B.(\pi r^2)cos 0\textdegree\\=1.9\pi\times 0.0925^2 \times cos 0\textdegree\\=5.107\times10^-^2 \ Tm^2[/tex]
we can now calculate the induced emf, [tex]\frac{\phi}{\bigtriangleup t}[/tex]:
[tex]\frac{\phi}{\bigtriangleup t}=\frac{5.107\times 10^-^2}{0.15}\\=3.405\times 10^-^1V[/tex]
Hence, the induced emf of the loop is 0.3405V
