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Coil 1 hasL1=25mHand N1=100turns. Coil 2 has L2=40mHand N2=200turns. The coils are fixed in place; their mutual inductance M is M=3.0mH. A 6.0mAcurrent in coil 1 is changing at the rate of 4.0 A/s. (a) What magnetic flux φ12links coil 1? (b) what self-induced emf appears in that coil? (c) What magnetic flux φ21links coil 2? (d) what mutually induced emf appears in that coil?

Short Answer

Expert verified

a)ϕ12=1.5×10-6Wb

b) ε=0.1V

c)ϕ21=9×10-8Wb

d)ε=0.012V

Step by step solution

01

Given

L1=25mH=0.025HN1=100L2=40mH=0.040HN2=200M=3.0mH=0.003HI=6.0mA=0.006Adi/dt=4.0A/s

02

Understanding the concept

We have the formula for the magnetic flux induced in the coil, the self-induced emf and the mutually induced emf. So, using these equations and the given data, we can solve this problem.

Formula:

ϕ=LINε=Ldidtϕ=MINε=Mdidt

03

(a) Calculate magnetic flux φ12 

We have,

ϕ=LIN

Magnetic flux ϕ12linking to coil 1 will be

ϕ12=L1IN1ϕ12=0.025×0.006100ϕ12=1.5×10-6Wb

04

(b) Calculate self-induced emf appeared in the coil 

We have the self – induced emf

ε=Ldidtε=0.025×4.0ε=0.1V

05

(c) Calculate magnetic flux φ21 

The flux can be written in terms of the mutual inductance, the current and the number of turns.

ϕ=MIN

Magnetic flux ϕ21linking to coil 2, will be denoted asϕ21.

ϕ21=0.003×0.006200ϕ21=9×10-8Wb

06

(d) Calculate mutually induced emf appears in the coil 

EMF induced can be written in terms of the mutual inductance and the rate of change of the current.

ε=Mdidtε=0.003×4.0ε=0.012V

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