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4 years ago

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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?

1 answer:

KIM [24]4 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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