An electric generator has an 18 -cm-diameter, 120 -turn coil that rotates at 60 $\mathrm{Hz}$in a uniform magnetic field that is perpendicular to the rotation axis. What magnetic field strength is needed to generate a peak voltage of 170$\mathrm{V}$?

The magnetic field lines has been the portion of magnetization which thus goes through one loop of across By This magnet is calculated on the basis:

${\mathrm{\Phi }}_m=BA$

The magnet is given by if the magnetism makes an equal with axis.

${\mathrm{\Phi }}_{\mathrm{m}}=NBA\cos \omega t=NBA\cos \left(2\pi ft\right)$

The rotation's pitch is represented by the sign f. Whereas a coil's diameter is d = 18 cm, the length equals $A=\pi {\left(\frac{d}{2}\right)}^2=\pi {\left(\frac{0.18\mathrm{m}}{2}\right)}^2=25.45\times {10}^{-3}{\mathrm{m}}^2$

From Faradays, the mediated is the modification of the magnetization in there in the spiral and this is provided by the type$\epsilon =-\frac{d{\mathrm{\Phi }}_{\mathrm{m}}}{dt}=-\frac{dNBA\cos \left(2\pi ft\right)}{dt}$

$=NBA\frac{d\cos \left(2\pi ft\right)}{dt}$

$=2\pi fNBA(\sin 2\pi ft)$

Because the magnetosphere is diagonally to the axis of rotation, the word "perpendicular magnetic field" is used.

$\sin 2\pi ft=\sin {90}^{\circ }=1$. We solve equation

$B=\frac{\epsilon }{2\pi fNA}$

The values $\epsilon ,f,N\text{and}A$

$B=\frac{\epsilon }{2\pi fNA}$

$=\frac{170\mathrm{V}}{2\pi \left(60\mathrm{Hz}\right)\left(120\right)\left(25.45\times {10}^{-3}{\mathrm{m}}^2\right)}$

$=0.15\mathrm{T}$