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Inessa05 [86]
3 years ago
15

A 0.110 m radius, 506-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a

uniform magnetic field. (This is 60 rev/s.) Find the magnetic field strength needed to induce an average emf of 10,000 V. 519.893 Incorrect: Your answer is incorrect.
Physics
1 answer:
8090 [49]3 years ago
3 0

Answer:

Explanation:

Expression for emf induced = 1 /2 B n r²ω where B is magnetic field , r is radius of the coil and ω is angular velocity of rotation , n is no of turn .

ω = 2πf = 2π x 60 = 120 π   ; where f is frequency of rotation .

emf induced =  1 /2 B r²x 120π x n

Putting the given values

10000 = .5 x B x .11² x 120 x 3.14 x 506

B = 8.67 T .

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8563732.58906 Pa

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P(V-Nb)=NkT\\\Rightarrow P=\dfrac{NkT}{V-Nb}\\\Rightarrow P=\dfrac{3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P=8563732.58906\ Pa

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Work done is given by

dw=Pdv\\\Rightarrow W=\int_{v_1}^{v_2}\dfrac{NkT}{V-Nb}dv\\\Rightarrow W=NkTln\dfrac{V_2-Nb}{V_1-Nb}\\\Rightarrow W=3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300ln\dfrac{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow W=5708.00923\ J

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