Two coils are wound around the same cylindrical form. When the current in the fi

Question

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.250 A/s , the induced emf in the second coil has a magnitude of 1.65×103 V .

Part A

What is the mutual inductance of the pair of coils?

Part B

If the second coil has 23 turns, what is the flux through each turn when the current in the first coil equals 1.25 A ?

Part C

If the current in the second coil increases at a rate of 0.355 A/s , what is the magnitude of the induced emf in the first coil?

Explanation / Answer

Let E be the induced emf,

M be the mutual inductance,

I be the current in the first coil,

N2 be the no. of turns in second coil

A)

Then, we have, E = M.dI/dT => M = E/(dI/dT) = (1.65*10^-3)/0.250 = 6.6*10^-3 H

B)

Flux through each turn = M.I/N2 = 6.6*10^-3*1.25/23 = 3.59*10^-4Wb

C)

E = M.dI/dT => E = 6.6*10^-3*0.355 = 2.343*10^-3V-----ans

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