ENE/EIE 325 Electromagnetic Fields and Waves Lecture 11 Magnetic Force, Magnetic Boundary Conditions, and Inductance

Force on a Moving Charge The forces exerted on a point charge by electric and magnetic fields are: E + Q Fe = QE B v X Fm Fe Fm = Q (v x B) in the direction of E into the screen charge can be stationary or moving charge is moving at velocity v

Cross product and right hand rule Let “a” represent the velocity “ ” and “b” represent the magnetic flux density “ ”, a  b = whose direction is represented by your thumb. then points in the direction of your thumb. https://figures.boundless.com/17800/full/right-20hand-20rule.png

Lorentz Force Law Generally, with both electric and magnetic fields present, we have both forces: E + B v Q The total force on the moving charge is then the sum of the two, or This is the Lorentz Force Law (sometimes called the “fifth Maxwell equation”) The electric field will, in this case, accelerate the charge in the direction of E, making it cross the B field lines in the perpendicular sense; this gives a magnetic force component that is out of the screen

Motion of the electric charge in a steady magnetic field Click at the image

The effect is called “Hall Effect”. Hall voltage is created by the separation of charges due to the magnetic force. The effect is called “Hall Effect”. Assumption: steady magnetic field (a) positive charges are moving inward and (b) negative charges moving outward in (b). The two cases can be distinguished by oppositely directed Hall voltages, as shown. Hayt, Buck, “Engineering Electromagnetics, 8th ed. pg. 251

Force on a Differential Current Element B dL J Consider a small segment (length dL) of current in the form of a volume current density J, suspended in a magnetic field, B. The current element has volume dv. We know that current density is volume charge density moving at velocity v: ..and we can write the differential force on a differential charge, dQ: where Therefore: ..so that finally: dv

Other Expressions for Differential Force For volume current, surface current, or filament current, we have the appropriate expressions for differential current: The corresponding expressions for differential force within magnetic field B are: volume current density (three dimensions) surface current density (two dimensions) filament current of length dL (one dimension)

Evaluating the Total Force In three or two dimensions, the net force is found by integrating over the volume or surface that the current occupies For a filament current, the total force will be the integral of the differential force, generally taken over the closed path that comprises the current: For a straight filament of length L, having uniform current, and within a uniform field, this becomes: which in turn reduces to

Example: Force on a Square Current Loop The magnetic field arising from the straight wire, evaluated in the plane of the loop, is: The B field is then: The total force on the loop is then found by evaluating: over all four loop segments

Example: Continued The integral becomes: A B C D

Magnetic boundary conditions (1) Gauss’s law for magnetostatics

Magnetic boundary conditions (2) Use Ampere’s circuital law or

Ex1 From the interface shown, given mT, determine and the angle that it makes with the interface.

Ex2 The interface between two magnetic materials is defined by the equation shown. Given H1 = 30 A/m, determine the following a)

b) c) d)

Ex3 Let the permeability be 5 H/m in region A where z > 0, and 7 H/m in region B where z < 0. If there is a surface current density A/m at z = 0, and if mT, find a) b)

c)

Potential energy of magnetic materials J/m3.

Duality of magnetostatics and electrostatics Electrostatics Magnetostatics V = IR Vm =   = Note: is magnetomotive force (mmf) [A]; is the magnetic flux,  is “reluctance” [Amperes x turns/Weber], is the cross-section area

Inductance Inductance (or self-inductance) is defined as the ratio of the total flux linkages to the current which they link, where L = inductance (Henry or 1 Weber-turn/A)  = total magnetic flux linking each of the turn (Weber) I = current flows in the N-turn coil (A) N = number of turns (turn)

Inductance is determined by flux linkage between turns Consider the toroid with N turns packed closely, Flux linkage = the product of the number of turns N and the flux linking each of them = N  I is the current flows in the N-turn coil. N       N Hayt, Buck, “Engineering Electromagnetics, 8th ed. pg. 265

Inductor Inductor is a passive electronic component that stores energy in the form of a magnetic field by using its property of inductance. http://www.electronicsandyou.com/electronics-basics/inductors.jpg

How does the inductor work? Click at the image

Inductor is a basic circuit component We assume the ideal switch. The inductor is used in the electrical circuit to gradually change the current flow. The coil will store or release energy in its magnetic field as rapidly as necessary to oppose any change in current. http://static.ddmcdn.com/gif/red-light-camera-loop.gif

Traffic light sensor The inductor is buried under the street. A car acts as a magnetic core that changes the inductance. The change in inductance triggers the traffic light. http://static.ddmcdn.com/gif/red-light-camera-loop.gif

A procedure for finding the inductance 1. Assume a current I in the conductor 2. Determine using the law of Biot-Savart, or Ampere’s circuital law if there is sufficient symmetry. 3. Calculate the total flux  linking all the loops. 4. Multiply the total flux by the number of loops to get the flux linkage. 5. Divide the flux linkage by I to get the inductance. The assumed current will be divided out.

Example: Inductance calculation of the toroid using Ampère’s law Let’s assume a toroid with the air core. Click at the image   From and   where r = the radius of the toroid http://www.youtube.com/watch?v=jdsUQs9w0uw&noredirect=1

Example: Inductance calculation of the toroid using Ampère’s law   r From and   If the dimensions of the cross section are small compared with the mean radius of the toroid r (r0 << r) , then the total flux is   r0   cross section http://www.youtube.com/watch?v=jdsUQs9w0uw&noredirect=1

Example: Inductance calculation of the toroid using Ampère’s law     H http://www.youtube.com/watch?v=jdsUQs9w0uw&noredirect=1

Inductance for a coaxial cable We can start by assuming a current going in the +az direction on the inner conductor. We can easily find the field between the conductors using Ampere’s circuital law. For (a < r < b ):

Inductance for a coaxial cable To find the flux linkage, we need to know the number of loops being linked by the flux. Notice in the figure that the innner and outer conductors are shown connected at the ends. These connections are considered to be a long distance from where we are calculating the inductance. It is easy to see that there is only one loop of current. The inductance per unit length is simply

Inductance for a solenoid with N turns Assuming that a current I is going into one end of the conductor. We found inside a solenoid. Here, we assume that h >> a, so we can neglect any edge effect and drop-off of the field. Then, within the r core we have where  = ro. The cross-sectional area of one loop in the solenoid is a2 so the total flux through a loop is

Inductance for a solenoid with N turns This flux is linked to the current N times, so the flux linkage is Finally, we divide out the current to find the inductance,

Suppose the toroid has a small spacing between turns, the inductance can then be found by sum of flux of each turn

More about inductance The definition of the inductance can be written in the form of magnetic energy as The current inside conductor creates the magnetic flux inside the material texture. This flux causes an internal inductance which combines with the external inductance to get the total inductance. Normally, the internal inductance can be neglected due to its small value compared to the external one.

Ex4 Calculate the inductance for the following configuration: a) A coaxial cable with the length l = 10 m, the inner radius a = 1 mm, and the outer radius b = 4 mm. The inserted magnetic material has r = 18 and r = 80.

b) A toroid with a number of turns N = 5000 turns with in = 3 cm, out = 5 cm, and the length l = 1.5 cm. The inserted material has r = 6. c) A solenoid has the radius r = 2 cm, the length l = 8 cm, and N = 900 turns. The inserted material has r = 100.

Mutual inductance When two or more sets of coils are magnetically linked together by a common magnetic flux they are said to have the property of Mutual Inductance. Mutual inductance is the basic operating principal of the transformer, motors, generators and any other electrical component that interacts with another magnetic field. http://www.electronics-tutorials.ws/inductor/mutual-inductance.html

Mathematical form of mutual inductance Mutual inductance (M) is mathematically defined as   where M12 = mutual inductance (H) 12 = the flux produced by coil L1 which links the path of the coil L2 I 1 = current applied to coil L1 (A) N2 = number of turns of coil L2 (turn)

M12 = M21 Likewise, the flux linking coil L1 when a current flows around coil L2 is exactly the same as the flux linking coil L2 when the same current flows around coil L1. Then the mutual inductance of coil one with respect of coil two is defined as M21. We can write the mutual inductance between the two coils as: M12 = M21 = M.