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Calculate the peak voltage (in Volts) of a generator that rotates its 200-turn, 0.100 m diameter coil at 3.5 x 103 rpm in a 0.7 T field. You should round your answer to the nearest integer, do not include unit.

Respuesta :

Angular velocity of the coil :

[tex]\omega=\dfrac{2\pi n}{60}[/tex]

[tex]\omega=\dfrac{2\times \pi\times 3500}{60}\\\\\omega =366.52\ rad/s[/tex]

Now,

Peak voltage is given by :

[tex]V=NAB\omega\\\\V= 200\times \dfrac{\pi (0.1)^2}{4}\times 0.7\times 366.52\\\\V=403.01\ V[/tex]

Hence, this is the required solution.

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