Angular velocity of the coil :
[tex]\omega=\dfrac{2\pi n}{60}[/tex]
[tex]\omega=\dfrac{2\times \pi\times 3500}{60}\\\\\omega =366.52\ rad/s[/tex]
Now,
Peak voltage is given by :
[tex]V=NAB\omega\\\\V= 200\times \dfrac{\pi (0.1)^2}{4}\times 0.7\times 366.52\\\\V=403.01\ V[/tex]
Hence, this is the required solution.
