Question

The coil rotates in a 0.15 T magnetic field at a speed of 50 revolutions per second. The area of ​​the coil is 1600 cm². How many turns of winding are needed to obtain an AC voltage with a frequency of 50 Hz and a peak voltage of 320 V?

Answer 1

Gives

$B=0.15T\\n=50Hz\\A=1600\times10^{-4}m^2\\V=320V\\n=50Hz$

We know that

$V=NBAw$

$$N=\frac{V}{BAw}=\frac{320}{0.15\times1600\times10^{-4}\times2\times3.14\times50}=42$$

N=42