MUTUAL INDUCTANCE

Suppose we hook up an AC generator to a solenoid so that the wire in the solenoid carries AC. Call this solenoid the primary coil. Next place a second solenoid connected to an AC voltmeter near the primary coil so that it is coaxial with the primary coil. Call this second solenoid the secondary coil. As shown in figure.

The alternating current in the primary coil produces an alternating magnetic field whose lines of flux link the secondary coil (like thread passing through the eye of a needle). Hence the secondary coil encloses a changing magnetic field. By Faraday’s law of induction this changing magnetic flux induces an emf in the secondary coil. This effect in which changing current in one circuit induces an emf in another circuit is called mutual induction.

Let the primary coil have N1 turns and the secondary coil have N₂ turns. Assume that the same amount of magnetic flux F₂from the primary coil links each turn of the secondary coil. The net flux linking the secondary coil is then N₂F₂. This net flux is proportional to the magnetic field, which, in turn, is proportional to the current I1 in the primary coil. Thus we can write N₂F₂µI₁. This proportionality can be turned into an equation by introducing a constant. Call this constant M, the mutual inductance of the two coils:

The unit of inductance is WB/A=Henry (H) named after Joseph Henry.

The emf induced in the secondary coil may now be calculated using Faraday’s law:

The above formula is the emf due to mutual induction.

Example

The apparatus used in Experiment EM-11B consists of two coaxial solenoids. A solenoid is essentially just a coil of wire. For a long, tightly-wound solenoid of n turns per unit length carrying current I the magnetic field over its cross-section is nearly constant and given by

. Assume that the two solenoids have the same cross-sectional area A. Find a formula for the mutual inductance of the solenoids.

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