Our values are,
[tex]B=1*10^{-5}T[/tex]
[tex]N=400[/tex]
[tex]r = 2.0cm = 0.02 m[/tex]
Number of rotations per second = 6
We begin calculating the cross-section Area, which is given by,
[tex]A= \pi r^2[/tex]
[tex]A= \pi (0.02)^2[/tex]
[tex]A= 1.2566*10^{-3} m^2[/tex]
We can calculate the angular speed through the number of rotations per second,
[tex]\omega = 6 rev[/tex]
[tex]\omega = 6(\frac{2}{pi})[/tex]
[tex]\omega =37.699 rad/s[/tex]
With this velocity we can now calculate the maximum emf,
[tex]E= BAN\omega[/tex]
[tex]E = (4*10^{-5})(1.2566*10^{-3})(400)(37.699)[/tex]
[tex]E= 7.579*10^{-4} V[/tex]
[tex]E= 7.6*10^{-4} V[/tex]
