ENE 325 Electromagnetic Fields and Waves Lecture 4 Magnetostatics

Introduction (1) source of the steady magnetic field may be a permanent magnet, and electric field changing linearly with time or a direct current. a schematic view of a bar magnet showing the magnetic field. Magnetic flux lines begin and terminate at the same location, more like circulation.

Introduction (2) Magnetic north and south poles are always together.

Introduction (3) Oersted’s experiment shows that current produces magnetic fields that loop around the conductor. The field grows weaker as one compass moves away from the source of the current.

Bio-Savart law (1) Bio-Savart law (1) The law of Bio-Savart states that at any point P the magnitude of the magnetic field intensity produced by the differential element is proportional to the product of the current, the magnitude of the differential length, and the sine of the angle lying between the filament and a line connecting the filament to the point P at which the filed is desired. The magnitude of the magnetic field intensity is inversely proportional to the square of the distance from the differential element to the point P.

Bio-Savart law (2) Bio-Savart law (2) The direction of the magnetic field intensity is normal to the plane containing the differential filament and the line drawn from the filament to the point P. Bio-Savart law is a method to determine the magnetic field intensity. It is an analogy to Coulomb’s law of Electrostatics.

Bio-Savart law (3) Bio-Savart law (3) from this picture: Total field A/m

Magnetic field intensity resulting from an infinite length line of current (1) Pick an observation point P located on  axis. The current The vector from the source to the test point is a unit vector

Magnetic field intensity resulting from an infinite length line of current (2) then From a table of Integral, then

Magnetic field intensity resulting from a ring of current (1) A ring is located on z = 0 plane with the radius a. The observation point is at z = h.

Magnetic field intensity resulting from a ring of current (2) A unit vector

Magnetic field intensity resulting from a ring of current (3) Consider a symmetry  components are cancelled out due to symmetry of two segments on the opposite sides of the ring. Therefore from we have

Magnetic field intensity resulting from a ring of current (4) then We finally get

Magnetic field intensity resulting from a rectangular loop of current (1) Find the magnetic field intensity at the origin. By symmetry, there will be equal magnetic field intensity at each half width (w/2).

Magnetic field intensity resulting from a rectangular loop of current (2) Consider 0  x  w/2, y = -w/2 A unit vector We have

Magnetic field intensity resulting from a rectangular loop of current (3) Then the total magnetic field at the origin is Look up the table of integral, we find then A/m.

Bio-Savart law in different forms We can express Bio-Savart law in terms of surface and volume current densities by replacing with and : where K = surface current density (A/m) I = K x width of the current sheet and

Right hand rule The method to determine the result of cross product (x).

Amp é re’s circuital law Analogy to Gauss’s law Use for magnetostatic’s problems with sufficient symmetry. Ampere’s circuital law – the integration of around any closed path is equal to the net current enclosed by that path. To find, choose the proper Amperian path that is everywhere either tangential or normal to and over which is constant. A

Use Ampere’s circuital law to determine from the infinite line of current. From then A/m.

Magnetic field of the uniform sheet of current (1) Create path a-b-c-d and perform the integration along the path.

Magnetic field of the uniform sheet of current (2) From divide the sheet into small line segments along x-axis, by symmetry H z is cancelled.. Because of the symmetry, the magnetic field intensity on one side of the current sheet is the negative of that on the other.

where is a unit vector normal to the current sheet. Magnetic field of the uniform sheet of current (3). Above the sheet, (z > 0) and (z < 0) or we can write A/m

Magnetic field inside the solenoid From

Magnetic field inside the toroid that has a circular cross section (1)

Magnetic field inside the toroid that has a circular cross section (2) From

Ex1 Determine at point P (0.01, 0, 0) m from two current filaments as shown.

Ex2 Determine for the coaxial cable that has a inner radius a = 3 mm, b = 9 mm, and c = 12 mm. Given I = 0.8 A. a) at  < a

b) at a <  < b c) at b <  < c d) at  > c

Ex3 Determine at point (10, 0, 0) mm resulted from three current sheets: K 1 = 1.5  A/m at x = 6 mm, K 2 = -3  A/m at x = 9 mm, and K 3 = 1.5  A/m at x = 12 mm.