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,

$E=m\frac{dI}{dT}$

Where E is the induced emf

I the current in the first coil

M the mutual inductance

Solving for a)

$M=\frac{E}{\frac{dI}{dT}}\\M=\frac{1.6*10^{-3}}{0.245}=6.53*10^{-3}H$

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

$\Phi=\frac{MI}{N}$

Where N is the number of turns in the second coil

$\Phi=\frac{6.53*10^{-3}*1.25}{22}=3.71*10^{-4}Wb$