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SVETLANKA909090 [29]
3 years ago
12

Calculate the peak voltage of a generator that rotates its 200-turn, 0.100 m diameter coil at 3600 rpm in a 0.800 T field.

Physics
1 answer:
m_a_m_a [10]3 years ago
6 0

Answer:

The peak voltage is 473.86 V

Explanation:

Given;

number of turns of generator, N = 200  turn

diameter of the coil, d = 0.1 m

radius of the coil, r = 0.05 m

Magnitude of the magnetic field, B = 0.8 T

angular frequency, ω = 3600 rpm

angular frequency, ω (rad/s) = \frac{2\pi}{60} *3600\ rpm = 377.04 \ rad/s

The peak voltage is given by;

E = NBAω

Where;

A is the area of the coil = πr² =  π (0.05)² = 7.855 x 10⁻³ m²

E = (200)(0.8)(7.855 x 10⁻³)(377.04)

E = 473.86 V

Therefore, the peak voltage is 473.86 V

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We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

Q=\frac{E}{Nr^2}

       

From the question we are told

Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e

<h3>An Expression for the <em>magnitude </em>of charge moved</h3>

Generally the equation for the  <em>magnitude </em>of charge moved, Q   is mathematically given as

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An expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be

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