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valentinak56 [21]
2 years ago
6

An electric generator consists of a circular coil of wire of radius 4.0×10−2 m , with 20 turns. The coil is located between the

poles of a permanent magnet in a uniform magnetic field, of magnitude 5.0×10−2 T . The B field is orientated perpendicular to the axis of rotation. The ends of the coil are connected via sliding contacts across a resistor of resistance 1.5 Ω. The peak current measured through the resistor is 3.0×10−3 A. What is the angular frequency ω at which the coil is rotating? in rad/s to two sig figs.
Physics
1 answer:
drek231 [11]2 years ago
8 0

Answer:

The angular frequency at which the coil is rotating is 17.9rad/s

Explanation:

To solve the exercise it is necessary to take into account the concepts related to the magnetic field in a wire, Farada's law and Ohm's Law.

The rotational induced voltage is defined by

V = NBA\omega

Where,

N = Number of loops

A = Cross-sectional area

\omega =angular velocity

B = Magnetic Field

For Ohm's law we have,

V = IR

Where,

I= Current

V = Voltage

R = Resistance

Equation both equations,

IR = NBA\omega

\omega = \frac{IR}{NBA}

Our values are given as,

N = 20

B =5*10^{-2}T

R = 1.5\Omega

I = 3*10^{-3}A

A = \pi r^2= \pi (4*10^{-2})^2

Replacing the values we have,

\omega = \frac{(3*10^{-3})(1.5)}{(20)(5*10^{-2})( \pi (4*10^{-2})^2)}

\omega = 17.904rad/s

Therefore the angular frequency at which the coil is rotating is 17.9rad/s

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Answer:

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<h3>How to get the position equation of the particle?</h3>

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$v(t)=\sin (\omega t) * \cos ^2(\omega t)

To get the position equation we just need to integrate the above equation:

$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t

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Replacing that in our integral we get:

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Then we have:

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To learn more about motion equations, refer to:

brainly.com/question/19365526

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