Equation (31.9) says that vab=Ldi>dt(see Fig. 31.8a). Using Faraday’s law, explain why point a is at higher potential than point b when i is in the direction shown in Fig. 31.8a and is increasing in magnitude. When i is counterclockwise and decreasing in magnitude, isvab=Ldi>dt still correct, or should it be vab=-Ldi>dt? Isvab=Ldi>dt still correct when i is clockwise and increasing or decreasing in magnitude? Explain.

Short Answer

Expert verified

Yes, the equationvab=Ldi>dt is correct when current I is flowing in clockwise or counterclockwise increasing or decreasing.

Step by step solution

01

State Faraday’s law

Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced. The induced emf in a coil is equal to the rate of change of flux linkage.

ε=-Ldidtwhere εis induced emf, Lis inductance.

02

Apply Faraday’s law

When variable current (i) passes through inductor, an emf is induces according to faraday’s law

ε=-Ldidt.

The point a is at higher potential than b because as the current increasesdidt is positive and induced emf is opposite to flow of current making a at higher potential.

vab=Ldidt.

vab=Ldidtthis expression is also correct when I is flowing counterclockwise and decreasing in magnitude becausedidt will be negative but sincevab will be in same direction as in current to support its decrement (as per faraday’s law).

By the same reason,vab=Ldidt is valid when current is either increasing or decreasing in clockwise direction.

Thus, point a is at higher potential than point b because current increases making thedidt positive and induced emf directed opposite to current. Also,vab=Ldidt is valid when current is flowing counterclockwise or clockwise in either increasing or decreasing manner.

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