Respuesta :
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) [tex]L = \dfrac{N \phi}{I}[/tex]
[tex]L = \dfrac{610\times 90 \times 10^{-6}}{2.80}[/tex]
[tex]L =19.6 mH[/tex]
b) inductance of A
[tex]L = \dfrac{N_A \phi_A}{I_A}[/tex]
[tex]L = \dfrac{430\times 300 \times 10^{-6}}{2.80}[/tex]
[tex]L =46 mH[/tex]
c) magnitude of the emf
[tex]\epsilon_B = -L_B\dfrac{dI}{dT}[/tex]
[tex]\epsilon_B = -(19.6\times 10^{-3})(0.5)[/tex]
[tex]\epsilon_B = -9.8\times 10^{-3}\ V[/tex]
[tex]\epsilon_B = -9.8\ mV[/tex]
