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Consider a generator that rotates its 200-turn, 0.15 m diameter coil at 3700 rpm in a 0.95 T magnetic field. What is the induced voltage in the coil?

Answer :

Final answer:

The peak induced voltage in the generator coil is calculated using Faraday's law of electromagnetic induction, with a final result of approximately 1311 volts.

Explanation:

Calculating Induced Voltage in a Generator

To calculate the peak voltage induced in a generator coil, we can use Faraday's law of electromagnetic induction. This law states that the induced electromotive force (emf) in a coil is equal to the negative change in magnetic flux through the coil times the number of turns in the coil. The formula to find the maximum induced emf (voltage) is given by:

e = NABω

where:

N is the number of turns in the coil,

A is the area of the coil,

B is the magnetic field strength, and

ω is the angular velocity of the coil in radians per second.

Given:

N = 200 turns,

Diameter of the coil, d = 0.15 m, so the radius r = d/2 = 0.075 m,

B = 0.95 T, and

Rotational speed = 3700 rpm.

To find the angular velocity:

ω = 2π × (Rotational speed in revolutions per second)

ω = 2π × (3700 / 60) rad/s = 2π × 61.67 rad/s

The area A of the coil is:

A = πr^2 = π × (0.075 m)^2

So, the peak induced voltage is:

e = NABω = 200 × π × (0.075 m)^2 × 0.95 T × 2π × 61.67 rad/s

e ≈ 200 × 0.0177 m^2 × 0.95 T × 387.6 rad/s

e ≈ 1311 V

Therefore, the peak induced voltage in the generator coil is approximately 1311 volts.

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