Edmund Li. Where is defined as the mutual inductance between and and has the SI units of Henries (H).

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1 INDUCTANCE MUTUAL INDUCTANCE If we consider two neighbouring closed loops and with bounding surfaces respectively then a current through will create a magnetic field which will link with as the flux passes through the surface. This mutual flux can be described as: Where is defined as the mutual inductance between and and has the SI units of Henries (H). If loop consists of turns, then the flux linkage can be denoted as: We can thus see that mutual inductance between two circuits is then the magnetic flux linkage with one circuit per unit current in the other. A more general definition of incremental mutual inductance is given by: By symmetry then, inductances on both loops will be identical:. If the medium is linear and isotropic, then the mutual SELF INDUCTANCE When the flux due to current links only with, itself, we refer to this phenomenon as self inductance the magnetic flux linkage per unit current in the loop itself: The self inductance of a loop depends on the geometrical shape and physical arrangement of the conductor, as well as on the permeability of the medium. With a linear medium, self inductance does not depend on the current in the loop or circuit nor on whether the loop is open or closed, or whether it is in the vicinity of another loop or circuit.

2 Thus, a conductor can be arranged in an appropriate shape to supply a certain amount of self inductance and is known as an inductor; a device which stores magnetic energy. From here, if we use inductance, we actually mean self inductance. To find the self inductance in a given inductor we must: 1. Choose a coordinate system 2. Assume a current I in the conducting wire 3. Find using Ampere s Law for symmetrical geometries or use Biot-Savart s Law 4. Find the flux 5. Find the flux linkage: 6. Find the inductance Assume that N turns of wire are tightly wound on a toroidal fram of a rectangular cross section with inner radius a and outer radius b and height h. Assuming that the permeability of the medium is, find the self inductance of the toroidal coil. 1. Clearly we will select the cylindrical coordinate system 2. Assume a current I flows in the wire 3. Using Ampere s Law: 4. Finding flux can be done by: 5. The flux linkage is then: 6. And the self inductance is:

3 An air coaxial transmission line has a solid inner conductor of radius a and a verythin outer conductor of inner radius b. Determine the inductance per unit length of the line. only has a - component due to cylindrical symmetry and is different inside the inner conductor and between the inner and outer conductors. a) Inside the conductor: The current through a cross sectional area radius r is given by:, so using Ampere s law: b) The outside radius: Now, applying the definition of flux to an annular ring of radius r we find the annular ring s contribution to the flux outside the ring: The logic behind such an integration is that the only time that the current will not always flow through the whole cross section (especially at high frequencies) due to the skin effect. The current flowing through the annulus ring is given by: Thus replacing with : The first term is known as the internal inductance which is due to flux linkage internal to the solid conductor. The second term is the external inductance due to the flux linkage between the inner and the outer conductor.

4 Calculate the internal and external inductances per unit length of a transmission line consisting of two long parallel conducting wires of radius a that carry currents in opposite directions. The axes of the wires are separated by a distance d, which is much larger than a. The internal inductance per unit length for each conductor is, thus the total internal inductance is To calculate the external inductance per unit length, we assume a current I. Since currents flow in opposite directions, only a y component exists between the two transmission lines. The flux linkage per unit length is then: For. Thus: Thus the total self inductance (sum of internal and external self inductance) per unit length is:

5 An iron ring of relative permeability 100 is wound with 2 coils of 100 and 400 turns. The cross section area of the ring core is 4, and the mean circumference is 50 cm. Calculate: a) The self inductance of each of the two coils b) The mutual inductance of the coils c) The total self inductance of the coils when connected in series with windings in the same sense d) The total self inductance of the coils when connected in series with windings in opposition a) But b) Now: c) If the wires are connected in the same sense, the flux is additive d) If the wires are connected in the opposite sense:

6 A steel ring of 50 cm mean length, and cross section area of is uniformly wound with a coil of 500 turns and a flux density of 1 T is produced by an exciting current of 4A. a) Calculate the coil inductance b) Calculate the exciting current necessary to maintain this flux density with a gap of 0.5 cm with is cut in the ring c) Find the inductance of the coil with the gap in the ring a) b) c)

7 ENERGY IN MAGNETIC FIELD An emf is induced in the loop that opposes the current. An amount of work must be done to overcome this induced emf. Given that the voltage across the inductance is: Then from the definition of work, which is the power times the change in time, the work required to increase the current from 0 to I is given by: Since, for linear media, then we can express work as: When there are two or more magnetic circuits, then the magnetic energy must also consider mutual coupling. Consider first, loop 1 having a current rising from 0 to and with, then: Similarly for loop 2: But since there are 2 magnetic circuits, there is also mutual coupling, some of the magnetic flux due to links with loop 1, giving rise to an induced emf that must be overcome by a voltage to keep constant at. The work involved is: in order Thus, the total work done in raising the currents in loops and from zero to and respectively is the sum of : We generalized this for N loops as: Where for linear media.

8 MAGNETIC ENERGY FROM FIELD QUANTITIES Given that and when N=1: Since: Then we can rewrite magnetic energy as: And even still, so: Where V is the volume of the loop. From the following vector identity: When the surface is large, A falls as 1/R and H falls as have:, so the second term is negligible. We thus We may define the magnetic density as: Such that:

9 A permanent magnet specimen is magnetized by a strong electromagnet created by a coil. After full magnetization, the coil current is reduced to zero and the magnet is then keepered by 3 keeper sections labeled A, B and C. All 3 keeper sections are infinitely permeable. The dimensions of the magnet and keep sections are. The cross sections of the magnet and the keepers to the left of the dotted line XY are uniform. The depths of the magnet specimen and the keepers are the same. The DC magnetization characteristic of the permanent magnet is also shown. a) With all 3 keeper sections in place, find the B field in the magnet, and the keeper section C. What are the stored energies in the magnet and the keepers? b) Keeper section C is removed and replaced by air. Find the B field and the magnet energy in the magnet. Where does this energy come from? a) from the graph. Since the cross sectional areas are the same, and the flux is constant, then. Since and. Thus b) Using Ampere s Law: Note that the keepers have so they are not included in the above equation. Thus: Since then: Since Finding the intersection of the loadline with the BH curve yields: This stored energy comes from the act to removing the keeper C.

10 By using stored magnetic energy, determine the inductance per unit length of an air coaxial transmission line that has solid inner conductor of radius and a very thin outer conductor of inner radius b. In the inner conductor: Now since we are dealing with per unit length, the volume and area is the same thing, so On the outside conductor: Since:

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