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A coil of 200 turns is rotated at 1700 r/min in a magnetic field having a uniform density of 0.05 t, the axis of rotation being at right angles to the direction of the flux. the mean area per turn is 40 cm2. calculate

a) the frequency
b) the period
c) the maximum value of the generated e.m.f.
d) the value of the generated emf when the coil has rotated through 30 degrees from the position of zero e.m.f.

1 Answer

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Final answer:

The frequency of the coil is 28.33 revolutions per second.

The period of the coil is 0.0352 seconds per revolution.

The maximum value of the generated emf is 557.9 volts

The value of the generated emf when the coil has rotated through 30 degrees is 80.0 volts.

Step-by-step explanation:

a) To calculate the frequency of the coil, we can convert the rotational speed from revolutions per minute to revolutions per second.

Since there are 60 seconds in a minute, the frequency can be calculated as 1700 revolutions per minute divided by 60 seconds per minute, which equals 28.33 revolutions per second.

b) To calculate the period, we can use the formula T = 1/f, where T is the period and f is the frequency.

Plugging in the calculated frequency of 28.33 revolutions per second, we get:

T = 1/28.33

= 0.0352 seconds per revolution.

c) To calculate the maximum value of the generated electromotive force (emf), we can use the formula:

E = N * B * A * ω, where E is the emf, N is the number of turns, B is the magnetic field density, A is the mean area per turn, and ω is the angular velocity.

Plugging in the given values, we get E = 200 * 0.05 * 40 * (2π * 1700/60)

= 557.9 volts.

d) To calculate the value of the generated emf when the coil has rotated through 30 degrees from the position of zero emf, we can use the formula E = N * B * A * ω * θ, where θ is the angle of rotation in radians.

Plugging in the values, we get :

E = 200 * 0.05 * 40 * (2π * 1700/60) * (30 * π/180)

= 80.0 volts.

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