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.[tex]M= 6.53\times10^{-3} H[/tex]
b.flux through each turn = Ф = [tex]4.08\times10^{-4} Wb[/tex]
c.magnitude of the induced emf in the first coil = e= [tex]2.38\times10^{-3} V[/tex]
Explanation:
a. rate of current changing = [tex]\frac{di}{dt}=[tex]M=\frac{1.60\times10^{-3} V}{0.240\frac{A}{s} }}[/tex][/tex]
 Induced emf in the coil =e= [tex]1.60\times10^{-3} V[/tex]
 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
   [tex]e=-M\frac{di}{dt}[/tex]
   [tex]M=\frac{e}{\frac{di}{dt} }[/tex]
   [tex]M=\frac{1.60\times10^{-3} }{-0.245}[/tex]
  [tex]M= 6.53\times10^{-3} H[/tex]
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 =  [tex]\frac{6.53\times10^{-3}\times1.25}{20}[/tex]
  flux through each turn = Ф = [tex]4.08\times10^{-4} Wb[/tex]
c.
  second coil increase at a rate = 0.365 A/s
 magnitude of the induced emf in the first coil =?
 using  [tex]e=-M\frac{di_{2} }{dt}[/tex]
      [tex]e= 6.53\times10^{-3} \times 0.365[/tex]
magnitude of the induced emf in the first coil = e= [tex]2.38\times10^{-3} V[/tex]
