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 A What is the mutual inductance of the pair of coils? MM = nothing H Request Answer Part B If the second coil has 22 turns, what is the flux through each turn when the current in the first coil equals 1.25 A ? ΦΦ = nothing Wb

Answer 1

To solve this problem we need to use the emf equation, that is,

`E=m(dI)/(dT)`

Where E is the induced emf

I the current in the first coil

M the mutual inductance

Solving for a)

`M=(E)/((dI)/(dT))\\M=(1.6*10^(-3))/(0.245)=6.53*10^(-3)H`

Solving for b) we need the FLux through each turn, that is

`\Phi=(MI)/(N)`

Where N is the number of turns in the second coil

`\Phi=(6.53*10^(-3)*1.25)/(22)=3.71*10^(-4)Wb`