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Varvara68 [4.7K]
3 years ago
14

Interactive Solution 22.43 provides one model for solving this problem. The maximum strength of the earth's magnetic field is ab

out 6.9 x 10-5 T near the south magnetic pole. Suppose we want to utilize this field with a rotating coil to generate 54.1-Hz ac electricity. What is the minimum number of turns (area per turn = 0.024 m2) that the coil must have to produce an rms voltage of 170 V?
Physics
1 answer:
laila [671]3 years ago
5 0

Answer:

427097

Explanation:

V = Voltage = 170 V

B = Magnetic field = 6.9\times 10^{-5}\ T

f = Frequency = 54.1 Hz

A = Area = 0.024\ m^2

Number of turns is given by

N=\dfrac{V\sqrt{2}}{BA2\pi f}\\\Rightarrow N=\dfrac{170\sqrt{2}}{6.9\times 10^{-5}\times 0.024\times 2\pi\times 54.1}\\\Rightarrow N=427096.933536\approx 427097\ turns

Number of turns is 427097

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K_i = 0 is the kinetic energy of the tool at the top (zero since it is at rest)

U_i = mgh is the gravitational potential energy of the tool at the top, where

m is the mass of the tool

g is the acceleration of gravity

h is the heigth of the tool

K_f = \frac{1}{2}mv^2 is the kinetic energy of the tool just before hitting the ground, where

v is the final speed of the tool

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Re-arranging the equation,

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g=9.8 m/s^2\\h=588 m

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v=\sqrt{2gh}=\sqrt{2(9.8)(588)}=107.4 m/s

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