Answer:[tex]410.90\times 10^{-6} V[/tex]
Explanation:
Given
[tex]A=5.5 mm^2[/tex]
[tex]n=919 turns/cm\approx 85400 turns/m[/tex]
[tex]\omega =226 rad/s[/tex]
According to the Faraday law of induction, induced emf is given by
[tex]E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}[/tex]
Magnetic field [tex]\phi _B=B\cdot A[/tex]
[tex]B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )[/tex]
[tex]B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)[/tex]
[tex]B=0.3305\sin (226t)[/tex]
[tex]\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}[/tex]
[tex]\phi _{B}=1.818\times 10^{-6}\sin (226t)[/tex]
[tex]E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}[/tex]
[tex]E=-1.818\times 10^{-6}\times 226\cos (226t)[/tex]
[tex]E=-410.90\times 10^{-6}\cos (226t)[/tex]
Amplitude of EMF induced[tex]=410.90\times 10^{-6} V[/tex]
