Answer:
4.19 V
Explanation:
Magnetic field flux and induced emf formula is:
[tex]\epsilon_{max}=NA\omega B_{max}[/tex]
Where
[tex]\epsilon_{max}[/tex] is the max emf (what we want)
N is the number of turns (N = 400)
A is the Area which is ([tex]\pi r^2[/tex])
[tex]\omega[/tex] is the angular velocity ( [tex]\omega = 2\pi f[/tex], f is frequency, f = 90)
[tex]B_{max}[/tex] is the max magnetic field value (given as 59 mT)
Finding A, Area:
[tex]A=\pi r ^2\\A= \pi (1)^2\\A=3.14[/tex]
In meters squared, it would be:
A = 3.14 * 10^-4 meter squared
Calculating angular velocity:
[tex]\omega = 2 \pi f\\\omega = 2 \pi (90)\\\omega = 565.49 rad/s[/tex]
Now, finding max EMF:
[tex]\epsilon_{max}=NA\omega B_{max}\\\epsilon_{max}=(400)(3.14*10^{-4})(565.49)(59*10^{-3})\\\epsilon_{max}=4.19[/tex]
So, the max value emf in the coil is around 4.19 V
