Question

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s , the induced emf in the second coil has a magnitude of 1.60×10−3 V .Part APart completeWhat is the mutual inductance of the pair of coils?MM= 6.53×10−3 HPrevious AnswersCorrectPart BPart completeIf the second coil has 20 turns, what is the flux through each turn when the current in the first coil equals 1.25 A ?ΦΦ= 4.08×10−4 WbPrevious AnswersCorrect

Answer 1

Answer:

Complete question:

c.If the current in the second coil increases at a rate of 0.365 A/s , what is the magnitude of the induced emf in the first coil?

a.$M= 6.53\times10^{-3} H$

b.flux through each turn = Ф = $4.08\times10^{-4} Wb$

c.magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

Explanation:

a. rate of current changing = $\frac{di}{dt}=[tex]M=\frac{1.60\times10^{-3} V}{0.240\frac{A}{s} }}$[/tex]

  Induced emf in the coil =e= $1.60\times10^{-3} V$

  For mutual inductance in which change in flux in one coil induces emf in the second coil given by the farmula based on farady law

     $e=-M\frac{di}{dt}$

     $M=\frac{e}{\frac{di}{dt} }$

     $M=\frac{1.60\times10^{-3} }{-0.245}$

   $M= 6.53\times10^{-3} H$

b.

  Flux through each turn=?

  Current in the first coil =1.25 A

   Number of turns = 20

       using   MI = NФ

     flux through each turn = Ф =  $\frac{6.53\times10^{-3}\times1.25}{20}$

   flux through each turn = Ф = $4.08\times10^{-4} Wb$

c.

   second coil increase at a rate = 0.365 A/s

  magnitude of the induced emf in the first coil =?

 using   $e=-M\frac{di_{2} }{dt}$

            $e= 6.53\times10^{-3} \times 0.365$

magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$