Answer:
The minimum no. of turns is [tex]3.126 \times 10^{5}[/tex]
Explanation:
Given:
Magnetic field [tex]B = 6.9 \times 10^{-5}[/tex] T
Frequency [tex]f = 84.5[/tex] Hz
Area of turn [tex]A = 0.021[/tex] [tex]m^{2}[/tex]
Voltage [tex]V_{rms} = 170[/tex] V
From the formula of induced emf,
[tex]V = NBA \omega[/tex]
Where [tex]\omega = 2\pi f[/tex] and [tex]V = \sqrt{2} V_{rms}[/tex]
So number of turn is,
[tex]N = \frac{\sqrt{2} V_{rms} }{AB2\pi f }[/tex]
[tex]N = \frac{\sqrt{2} \times 170 }{0.021 \times 6.9 \times 10^{-5} \times 6.28 \times 84.5 }[/tex]
[tex]N = 3.126 \times 10^{5}[/tex]
Therefore, the minimum no. of turns is [tex]3.126 \times 10^{5}[/tex]