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olga55 [171]
3 years ago
13

Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, have 430 and 610 turns, respectively. A

current of 2.80 A in solenoid A produces an average flux of 300
μ
Wb through each turn of A and a flux of 90.0
μ
Wb through each turn of B. (a) Calculate the mutual inductance of the two solenoids. (b) What is the inductance of A? (c) What is the magnitude of the emf that is induced in B when the current in A changes at the rate of 0.500 A/s?
Physics
1 answer:
KIM [24]3 years ago
7 0

Answer

given,

Two solenoids A and B

Number of turn

Na = 430 turns          Nb = 610 turns

Current = 2.80 A

Average flux through  A  = 300 μWb

Average of flux through B = 90.0  μ Wb

a) L = \dfrac{N \phi}{I}

   L = \dfrac{610\times 90 \times 10^{-6}}{2.80}

   L =19.6 mH

b) inductance of A

   L = \dfrac{N_A \phi_A}{I_A}

   L = \dfrac{430\times 300 \times 10^{-6}}{2.80}

   L =46 mH

c) magnitude of the emf

    \epsilon_B = -L_B\dfrac{dI}{dT}

    \epsilon_B = -(19.6\times 10^{-3})(0.5)

    \epsilon_B = -9.8\times 10^{-3}\ V

    \epsilon_B = -9.8\ mV

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The spectral density energy is related with the temperature and the wavelength (Planck’s law).

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The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

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A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

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Where \sigma is the Stefan-Boltzmann constant and T is the temperature.

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It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

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Before replacing all the values in equation (4), \lambda max (450 nm) will be express in meters:

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