Preparatory Notes and Examples for Exam #2

1. Preparatory Notes and Examples for Exam #2 EEE241: Fundamentals of Electromagnetic Theory Instructor: Dragica Vasileska

2. Electromagnetic Force The electromagnetic force is given by Lorentz Force Equation (After Dutch physicist Hendrik Antoon Lorentz (1853 – 1928)) The Lorentz force equation is quite useful in determining the paths charged particles will take as they move through electric and magnetic fields. If we also know the particle mass, m, the force is related to acceleration by the equation The first term in the Lorentz Force Equation represents the electric force Fe acting on a charge q within an electric field is given by. The electric force is in the direction of the electric field.

3. Magnetic Force The second term in the Lorentz Force Equation represents magnetic force Fm(N) on a moving charge q(C) is given by where the velocity of the charge is u (m/sec) within a field of magnetic flux density B (Wb/m2). The units are confirmed by using the equivalences Wb=(V)(sec) and J=(N)(m)=(C)(V). The magnetic force is at right angles to the magnetic field. The magnetic force requires that the charged particle be in motion. It should be noted that since the magnetic force acts in a direction normal to the particle velocity, the acceleration is normal to the velocity and the magnitude of the velocity vector is unaffected. Since the magnetic force is at right angles to the magnetic field, the work done by the magnetic field is given by

4. Magnetostatics – Biot-Savart’s Law Shortly following Oersted’s discovery that currents produce magnetic fields, Jean Baptiste Biot (1774-1862) and Felix Savart (1791-1841) arrived at a mathematical relation between the field and current. The Law of Biot-Savart is (A/m) To get the total field resulting from a current, you can sum the contributions from each segment by integrating (A/m) Note: The Biot-Savart law is analogous to the Coulomb’s law equation for the electric field resulting from a differential charge

5. H Magnetostatics – Ampere’s Circuital Law In electrostatics problems that featured a lot of symmetry we were able to apply Gauss’s Law to solve for the electric field intensity much more easily than applying Coulomb’s Law. Likewise, in magnetostatic problems with sufficient symmetry we can employ Ampere’s Circuital Law more easily than the Law of Biot-Savart. Ampere’s Circuital Law says that the integration of H around any closed path is equal to the net current enclosed by that path. The line integral of H around a closed path is termed the circulation of H.

6. Magnetostatics – Magnetic Flux Density The magnetic flux density, B, related to the magnetic field intensity in free space by where o is the free space permeability, given in units of henrys per meter, or The units of B are therefore (H)(A)/m2, but it is more instructive to write webers per meter squared, or Wb/m2, where Wb=(H)(A). But for brevity, and perhaps to honor a deserving scientist, a tesla , T, equivalent to a Wb/m2, is the standard unit adopted by the International System of Units. The amount of magnetic flux, , in webers, from magnetic field passing through a surface is found in a manner analogous to finding electric flux:

7. Closed Surface I5 I4 I3 I1 I2 Node Gauss’s Law and Kirchhoff’s Current Law Gauss’s Law: The net magnetic flux passing through a closed surface (Gaussian surface) must be zero Therefore, the algebraic sum of the currents entering any closed surface is zero. This is analogous to Kirchhoff’s Current Law (KCL)! Kirchhoff’s Current Law: The algebraic sum of the currents entering any node is zero.

8. Magnetostatic Boundary Conditions Special Case: If the surface current density K = 0, we get If K = 0 The tangential magnetic field intensity is continuous across the boundary when the surface current density is zero. Important Note: (or) We know that Using the above relation, we obtain Therefore, we can say that The tangential component of the magnetic flux density B is not continuous across the boundary.

9. Magnetostatic Boundary Conditions Gauss’s Law for Magnetostatic fields: To find the second boundary condition, we center a Gaussian pillbox across the interface as shown in Figure. We can shrink h such that the flux out of the side of the pillbox is negligible. Then we have Normal BC:

10. Magnetic Force Example: At a particular instant in time, in a region of space where E = 0 and B = 3ay Wb/m2, a 2 kg particle of charge 1 C moves with velocity 2ax m/sec. What is the particle’s acceleration due to the magnetic field? Given: q= 1 nC, m = 2 kg, u = 2 ax (m/sec), E = 0, B = 3 ay Wb/m2. Lorentz Force Equation Newtons’ Second Law Equating To calculate the units: Example: A 10. nC charge with velocity 100. m/sec in the z direction enters a region where the electric field intensity is 800. V/m ax and the magnetic flux density 12.0 Wb/m2ay. Determine the force vector acting on the charge. Given: q= 10 nC, u = 100 az (m/sec), E = 800 ax V/m, B = 12.0 ay Wb/m2.

11. Magnetic Force Example: A pair of parallel infinite length lines each carry current I = 2A in the same direction. Determine the magnitude of the force per unit length between the two lines if their separation distance is (a) 10 cm, (b)100 cm. Is the force repulsive or attractive? Magnetic force between two current elements when current flow is in the same direction Magnetic force per unit length Case (a) y = 10 cm Case (a) y = 10 cm

12. Magnetostatics – An Infinite Line current Example: Consider an infinite length line along the z-axis conducting current I in the +az direction. We want to find the magnetic field everywhere. We first inspect the symmetry and see that the field will be independent of z and  and only dependent on ρ. An infinite length line of current So we consider a point a distance r from the line along the ρ axis. The Biot-Savart Law IdL is simply Idzaz, and the vector from the source to the test point is The Biot-Savart Law becomes

13. Magnetostatics – An Infinite Line current Pulling the constants to the left of the integral and realizing that az x az = 0 and az x aρ = a, we have The integral can be evaluated using the formula given in Appendix D When the limit

14. Magnetostatics – An Infinite Line current An infinite length line of current Using We find the magnetic field intensity resulting from an infinite length line of current is Direction: The direction of the magnetic field can be found using the right hand rule. Magnitude: The magnitude of the magnetic field is inversely proportional to radial distance.

15. Magnetostatics – A Ring of Current Example: Let us now consider a ring of current with radius a lying in the x-y plane with a current I in the +az direction. The objective is to find an expression for the field at an arbitrary point a height h on the z-axis. The Biot-Savart Law The differential segment dL = ada The vector drawn from the source to the test point is Magnitude: Unit Vector: The biot-Savart Law can be written as

16. Magnetostatics – A Ring of Current We can further simplify this expression by considering the symmetry of the problem az components add A particular differential current element will give a field with an aρ component (from a x az) and an az component (from a x –aρ). Taking the field from a differential current element on the opposite side of the ring, it is apparent that the radial components cancel while the az components add. aρ Components Cancel H At h = 0, the center of the loop, this equation reduces to

17. Magnetostatics – A Solenoid A solenoid Solenoids are many turns of insulated wire coiled in the shape of a cylinder. Suppose the solenoid has a length h, a radius a, and is made up of N turns of current carrying wire. For tight wrapping, we can consider the solenoid to be made up of N loops of current. To find the magnetic field intensity from a single loop at a point P along the axis of the solenoid, from we have The differential amount of field resulting from a differential amount of current is given by The differential amount of current can be considered a function of the number of loops and the length of the solenoid as

18. Magnetostatics – A Solenoid Fixing the point P where the field is desired, z’ will range from –z to h-z, or This integral is found from Appendix D, leading to the solution At the very center of the solenoid (z = h/2), with the assumption that the length is considerably bigger than the loop radius (h >> a), the equation reduces to

19. Magnetostatics – Ampere’s Circuital Law Application to Current Sheet Example: Let us now use Ampere’s Circuital Law to find the magnetic field intensity resulting from an infinite extent sheet of current. Let us consider a current sheet with uniform current density K = Kxax in the z = 0 plane along with a rectangular Amperian Path of height h and width w. Amperian Path: In accordance with the right hand rule, where the thumb of the right hand points in the direction of the current and the fingers curl in the direction of the field, we’ll perform the circulation in the order Ampere’s Circuital Law

20. Magnetostatics – Ampere’s Circuital Law Application to Current Sheet From symmetry arguments, we know that there is no Hx component. Above the sheet H = Hy(-ay) and below the sheet H = Hyay. The current enclosed by the path is just Equating the above two terms gives A general equation for an infinite current sheet: where aN is a normal vector from the sheet current to the test point.

21. Magnetostatic Boundary Conditions Example: The magnetic field intensity is given as H1 = 6ax + 2ay + 3az (A/m) in a medium with r1 = 6000 that exists for z < 0. We want to find H2 in a medium with r2 = 3000 for z >0. Step (a) and (b): The first step is to break H1 into its normal component (a) and its tangential component (b). Step (c): With no current at the interface, the tangential component is the same on both sides of the boundary. Step (d): Next, we find BN1 by multiplying HN1 by the permeability in medium 1. Step (e): This normal component B is the same on both sides of the boundary. Step (f): Then we can find HN2 by dividing BN2 by the permeability of medium 2. Step (g): The last step is to sum the fields .

22. turns Example on Magnetic Circuits A coils of 200 turns is uniformly wound around a wooden ring with a mean circumference of 600 mm and area of cross-section of 500 mm2. If the current flowing into the coil is 4 A, Calculate (a) the magnetic field strength , (b) flux density dan (c) total flux N = 200 turns l = 600 x 10-3 m A = 500 x 10-6 m2 I = 4 A (a) H = NI/l = 200 x 4 / 600 x 10-3 = 1333 A (b) B = oH = 4 x 10-7 x 1333 = 0.001675 T = 1675 T (c) Total Flux  = BA = 1675 x 10-6 x 500 x 10-6 = 0.8375 Wb

23. Reluctance ( S ) Ohm‘s law I = V/R [A] Where I =current; V=voltage and R=resistance And the resistance can be relate to physical parameters as R =  l /A ohm Where =resistivity [ohm-meter], l= length in meter and A=area of cross-section [meter square] Analogy to the Ohm‘s law V=NI=H l I=F and R=S where

24. Magnetic circuit with different materials and For A: area of cross-section = a1 mean length = l1 absolute permeability = 1 For B: area of cross-section= a2 mean length= l2 absolute permeability= 2

25. Example on Magnetic Circuits A magnetic circuit made of silicon steel is arranged as in the Figure. The center limb has a cross-section area of 800mm2 and each of the side limbs has a cross-sectional area of 500mm2. Calculate the m.m.f required to produce a flux of 1mWb in the center limb, assuming the magnetic leakage to be negligible.

26. A B Looking at graph at B=1.25T mr =34000 Apply voltage law in loops A and B

27. Since the circuit is symmetry FA =FB In the center limb , the flux is 1mWb which is equal to 2 F Therefore F=0.5mWb A