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

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*10^(-3) H`

b.flux through each turn = Ф =
`4.08*10^(-4) Wb`

c.magnitude of the induced emf in the first coil = e=
`2.38*10^(-3) V`

Step-by-step explanation:

a. rate of current changing =
`(di)/(dt)=[tex]M=(1.60*10^(-3) V)/(0.240(A)/(s) )}`[/tex]

Induced emf in the coil =e=
`1.60*10^(-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(di)/(dt)`

`M=(e)/((di)/(dt) )`

`M=(1.60*10^(-3) )/(-0.245)`

`M= 6.53*10^(-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 =
`(6.53*10^(-3)*1.25)/(20)`

flux through each turn = Ф =
`4.08*10^(-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(di_(2) )/(dt)`

`e= 6.53*10^(-3) * 0.365`

magnitude of the induced emf in the first coil = e=
`2.38*10^(-3) V`