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A lecture slide on electromagnetic fields and waves

A lecture slide on electromagnetic fields and waves

- 1. Magnetostatics Force on a charged particle Q in a Electric field E Fe=QE Magnetic force Fm experienced by a charge Q moving with a velocity u in a magnetic field B is Fm=QuXB For a moving charge Q in the presence of both electric and magnetic fields, the total force on the charge is given by This is known as the Lorentz force equation for Electromagnetics
- 2. Fundamental Postulates of Magnetostatics The two fundamental postulate of magnetostatics that specify the divergence and and the curl of B in nonmagnetic media are Taking the volume integral of divergence equation and applying divergence theorem There are no magnetic flow sources, and the magnetic flux lines always close upon themselves. Integral Equation is also referred to as an expression for the law of conservation of magnetic flux because it states that the total outward magnetic flux through any closed surface is zero.
- 3. The integral form of the curl relation in the equation 2 can be obtained by integrating both sides over an open surface and applying Stokes's theorem. We have where the path C for the line integral is the contour bounding the surface S, and I is the total current through S. It is a form of Ampere's circuital law, which state that the circulation of the magnetic flux density in a nonmagnetic medium around any closed path is equal to µ0times the total current flowing through the surface bounded by the path Ampere’s Circuital Law
- 4. An infinitely long, straight, solid, nonmagnetic conductor with a circular cross-section of radius b carries a steady current I. Determine the magnetic flux density both inside and outside the conductor. If we align the conductor along the z-axis the magnetic flux density B will be -directed and will be constant along any circular path around the z-axis. z I Problem
- 5. Vector Magnetic Potential (A) · B = 0 (1), assure that B is solenoidal as consequence B can be expressed as the curl of another vector field A, such that .XA=0 (2) Comparing 1 and 2 we can write B=XA XB=0J Coulomb condition for divergence of A .A=0 Applying the condition we get This is known as vector Poisson’s equation
- 6. Thus, for Cartesian coordinates the Laplacian of a vector field A is another vector field whose components are the Laplacian (the divergence of the gradient) of the corresponding components of A Each of these three equation is mathematically the same as the scalar Poisson's equation.
- 7. Vector potential A relate to the magnetic flux through a given area S that is bounded by contour C in a simple way Thus, vector magnetic potential A does have physical significance in that its line integral around any closed path equal the total magnetic flux passing through the area enclosed by the path. Biot Savart Law For a thin wire with cross-sectional area S. dv’ equals S dl', and the current flow i entirely along the wire.
- 8. Primed indicate the source co-ordinate and unprimed indicate the space co-ordinate Since the unprimed and primed coordinates are independent, x dl' equals 0 dl’ (x’,y’,z’) P(x,y,z) aR R The distance R is measured from dl’ at (x', y’, z’) to the field point at (x, y, z). Thus we have -
- 9. where aR is the unit vector directed from the source point to the field point. Substituting in the equation This is known as Biot savart law for finding magnetic flux density from a current carrying element dB=
- 10. To determine the magnetic field intensity due to a straight current carrying filamentary conductor of finite length AB as in Figure. We assume that the conductor is along the z-axis with its upper and lower ends respectively subtending angles 2 and 1at P, the point at which H is to be determined. If we consider the contribution dH at P due to an element dl at(0,0,z), H=B/0 for nonmagnetic material
- 11. As a special case when the conductor is semi infinite (with respect to P) so that point A is now at (Origin (0,0,0)) while B is at (0,0, ); 1,=90°, 2=0°,and eq. becomes Another special case is when the conductor is infinite in length. For this case, point A is at (0,0,-) while B is at (0,0,); 1 =180°, and 2=0°
- 12. Find the magnetic flux density at a point on the axis of a circular loop of radius b that carries a direct current I. Problem Applying Biot-Savart law to the circular loop as shown in Fig. Again it is important to remember that R is the vector from the source element dl' to the field point P.
- 13. Because of cylindrical symmetry, it is easy to see that the ar component is cancelled by the contribution of the element located diametrically opposite to dl', Only consider the az component of this cross product and after integration At the center of the loop z=0 B=az
- 14. Find the magnetic flux density at a distant point of a small circular loop of radius b that carries a current I (a magnetic dipole) First Find vector magnetic potential then B where R 1 denotes the distance between the source element dl' at P' and the field point P(R,,/2) It is important to note that a at dl’ is not the same as a at point P. In fact, a, at P is ax Problem (The magnetic dipole) m aR mXaR=Ib2.1sin
- 15. For every Idl' there is another symmetrically located differential current element on the other side of the y-axis that will contribute an equal amount to A in the - ax direction but will cancel the contribution of I dl' in the ay direction The law of cosine applied to the triangle OPP' gives where R cos is the projection of R on the radius 0P', which is the same as the projection of OP" (OP"= Rsin) on OP'. Hence,
- 16. which yields The magnetic flux density B = x A. The expression of magnetic potential in terms of magnetic dipole moment z y x I b M=Isaz = Ib2 A-m2 a b I
- 17. A direct current I flows in an infinitely long wire of a radius 2(mm) along the z-axis. a) Obtain the vector magnetic potential A at r > 2 (mm) from the expression of B = xA . Choose the reference zero potential at wire surface. b) If I = 10 (A), determine from A the total amount of magnetic flux passing through a square loop specified by z = ± 0.3 (m) and y = 0.1 (m) and 0.7 (m). r0 = 2 mm 0.1 0.7 -0.3m 0.3 m y I B=XA A=azAz Ar=0, A=0 XA=-a r> 2 mm, so A is for outside the conductor For outside the conductor B=a (1) (2) z Comparing 1 and 2 𝜕𝐴 𝜕𝑟 = − 𝜇𝐼 2𝜋𝑟 After integration Az=- r=r0, A=0 C= A=az Equating 1 and 2 1 2 3 4
- 18. = For side 1 1= . . . dz dl=azdz 3=- . . . dz 2=0, 4=0 = 1+2+3+4
- 19. A very long straight conductor located along the z-axis carries a current I in the z-direction. Obtain an expression for the vector magnetic potential at a point in the bisecting plane of the conductor. What is the magnetic flus density at that point. A current carrying conductor extending in the z-direction from –L to L . The distance vector R of point P
- 20. MAGNETIZATION AND EQUIVALENT CURRENT DENSITIES ln the absence of an external magnetic field, the magnetic dipoles of the atom of most materials (except permanent magnets) have random orientations, resulting in no net magnetic moment. The application of an external magnetic field causes both an alignment of the magnetic moment of the spinning electrons and an induced magnetic moment due to a change in the orbital motion of electrons. To obtain a formula for determining the quantitative change in the magnetic flux density caused by the presence of a magnetic material Let mk is the magnetic dipole moment of an atom. If n is the number of atom per volume the magnetization vector M The magnetic dipole moment dm of an elemental volume dv' is dm = M dv' that will produce a vector magnetic potential
- 21. Magnetization surface current density Magnetization volume current density Magnetic-moment density M produces an internal flux density B; which is proportional to M. We may write
- 22. The magnetic field intensity H, such that When the magnetic properties of the medium are linear and isotropic, the magnetization is directly proportional to the magnetic field intensity Where m is a dimensionless quantity called magnetic susceptibility.
- 23. BOUNDARY CONDITIONS FOR MAGNETOSTATIC FIELDS From the divergence less nature of the B field The normal component of B is continuous across interface The tangential component of magnetic field is not continuous if there is a surface current along the interface.
- 24. INDUCTANCES AND INDUCTORS B1 is directly proportional lo l1 hence 12 is also proportional to l 1 In case C2 has N2 turn the flux linkage 12 due to 12 is
- 25. An air coaxial transmission line has a solid inner conductor of radius a and a very thin outer conductor of inner radius b. Determine the inductance per unit length of the line.
- 26. We first find the internal inductance L by considering the flux linkages due to the Inner conductor. From Figure, the flux leaving a differential shell of thickness d is
- 27. We now determine the external inductance Lext by considering the flux linkages between the inner and the outer conductor as in Figure For a differential shell of thickness d,
- 28. Boundary condition between two media Two different magnetic media
- 29. Problem: Since y-x- 2=0 is a plane, y-x<2 or y< x+2 is region1 in Figure. A Point in this region may be used to confirm this for example, the origin (0,0) is in this region since 0-0-2<0. If we let the surface of the plane be described by f(x,y)=y—x—2,a unit vector normal to the plane is given by
- 30. Two coaxial circular wires of radii a and b(b>a) are separated by distance h (h>a,b) as shown in Figure. Find the mutual inductance between the wires Problem Let current I1 flow in wire1.At an arbitrary point P on wire2, the magnetic vector potential due to wire1 is given by
- 31. A current generator i connected to the loop, which increases the current i1 from zero to I1. From physics we know that an electromotive force (emf) will be induced in the loop that opposes the current change. An amount of work must be done to overcome this induced emf. Let The work required is Now consider two closed loops C1 and C2 carrying current i1 and i2 respectively. The currents are initially zero and are to be increased to I1 and I2, respectively. To find the amount of work required, we first keep i2 =0 and increase i1 from zero to I1 This requires a work W1 in loop C1 Next we keep i1 at I1and increase i2from zero to I2 Because of mutual coupling, some of the magnetic flux due to i2 will link with loop C1, giving rise to an induced emf that must be overcome by a voltage v21 = ± L21 di2/dt in order to keep i1 constant at its value I1 • The work involved is Magnetic Stored Energy
- 32. The total amount of work done in raising the currents in loops C1 and C2 from zero to I1 and I2 respectively, is then the sum of W1, W21, and W2 which is the energy stored in the magnetic field of the two coupled current carrying loops. For a current I flowing in a single inductor with inductance L, the stored magnetic energy is Magnetic stored energy in terms of field quantities Wm= 𝐼= ∮ 𝐵𝑑𝑠 ∮ 𝐻. 𝑑𝑙 . = ∮ 𝐵. 𝐻𝑑𝑠𝑑𝑙 Wm= ∫ 𝐵. 𝐻𝑑𝑣 From Energy Formula
- 33. By using stored magnetic energy, determine the inductance per unit length of an air coaxial transmission line that has a solid inner conductor of radius a and a very thin outer conductor of inner radius b. Problem
- 34. Inside conductor Outside conductor Total Magnetic stored energy per unit length= Per unit length inductance:
- 35. Magnetic Force Let us consider an element of conductor dl with a cross -sectional area S. If there are N charge carriers (electrons) per unit volume moving with a velocity u in the direction of dl. then the magnetic force on the differential element is. according to u and dl have the same direction. Now. since - NeSlul equal the current in the conductor, The magnetic force on a complete (closed) circuit of contour C that carries a current I in a magnetic field B is then
- 36. When we have two circuits carrying currents I1 and I2, respectively, The situation is that of one current-carrying circuit in the magnetic field of the other. In the presence of the magnetic flux B12, which was caused by the current I1in C1 the force F12 on circuit C 2 can be written a From Biot Savart Law which is Ampere's law of force between two current-carrying circuits.
- 37. A rectangular loop carrying current I2 is placed parallel toan infinitely long filamentary Wire carrying current I1 As shown in Figure. Find the force experienced by the Loop Let the force on the loop be
- 38. Problem
- 39. The torque T (or mechanical moment of force) on the loop is the vector product of the force F and the moment arm r. N-m Let us apply this to a rectangular loop of length l and width w placed in a uniform Magnetic field B as shown in Figure. From this figure,we notice that dl is parallel to B along sides 12 and 34 of the loop and no force is exerted on those sides. Thus
- 40. where|F0 |=iBl because B is uniform View of figure, the torque on the loop T as magnetic dipole moment of the loop an is a unit normal Vector to the plane of the loop and its direction is determined by the right-hand rule fingers In the direction of current and thumb along an
- 41. The coil makes an angle 300 with the uniform magnetic flux density of 1.2 T as depicted in the figure. Determine the torque experienced by the coil if it carries a current of 50 A. (a) Circular coil immersed in the B field Side view of fig (a)
- 42. The side view of the coil indicating the direction of the dipole moment is shown in fig (b). The magnetic dipole moment lies in the xy plane and has a magnitude of m=NIA=200x50x10x10-4 =10 AT.m2

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