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AMPERE’S CIRCUITAL LAW (Cont’d) Expression for curl by applying Ampere’s Circuital Law might be too lengthy to derive, but it can be described as: H J The expression is also called the point form of Ampere’s Circuital Law, since it occurs at some particular point. 1 AMPERE’S CIRCUITAL LAW (Cont’d) The Ampere’s Circuital Law can be rewritten in terms of a current density, as: H dL J dS Use the point form of Ampere’s Circuital Law to replace J, yielding: H dL H dS This is known as Stoke’s Theorem. 2 3.3 MAGNETIC FLUX DENSITY In electrostatics, it is convenient to think in terms of electric flux intensity and electric flux density. So too in magnetostatics, where magnetic flux density, B is related to magnetic field intensity by: B H 0 r Where μ is the permeability with: 0 4 10 7 H m 3 MAGNETIC FLUX DENSITY (Cont’d) The amount of magnetic flux, φ in webers from magnetic field passing through a surface is found in a manner analogous to finding electric flux: B dS 4 MAGNETIC FLUX DENSITY (Cont’d) Fundamental features of magnetic fields: • The field lines form a closed loops. It’s different from electric field lines, where it starts on positive charge and terminates on negative charge Figure 3-26 (p. 125) Magnetic field lines form closed loops, sot he netflux through a Gaussian surface is zero. Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth Copyright © 2005 by John Wiley & Sons. All rights reserved. 5 MAGNETIC FLUX DENSITY (Cont’d) • The magnet cannot be divided in two parts, but it results in two magnets. The magnetic pole cannot be isolated. Figure 3-27 (p. 125) Dividing a magnet in two parts results in two magnets. You cannot isolate a magnetic pole. Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth Copyright © 2005 by John Wiley & Sons. All rights reserved. 6 MAGNETIC FLUX DENSITY (Cont’d) The net magnetic flux passing through a gaussian surface must be zero, to get Gauss’s Law for magnetic fields: B dS 0 By applying divergence theorem, the point form of Gauss’s Law for static magnetic fields: B 0 7 EXAMPLE 6 Find the flux crossing the portion of the plane φ=π/4 defined by 0.01m < r < 0.05m and 0 < z < 2m in free space. A current filament of 2.5A is along the z axis in the az direction. Try to sketch this! 8 SOLUTION TO EXAMPLE 6 The relation between B and H is: B 0 H 0 I 2 a To find flux crossing the portion, we need to use: B dS where dS is in the aφ direction. 9 SOLUTION TO EXAMPLE 6 (Cont’d) So, dS ddza Therefore, B dS 0 I a ddza z 0 0.01 2 2 0.05 20 I 0.05 6 ln 1.61 10 Wb 2 0.01 10 3.4 MAGNETIC FORCES Upon application of a magnetic field, the wire is deflected in a direction normal to both the field and the direction of current. (a) (b) 11 MAGNETIC FORCES (Cont’d) The force is actually acting on the individual charges moving in the conductor, given by: Fm qu B By the definition of electric field intensity, the electric force Fe acting on a charge q within an electric field is: Fe qE 12 MAGNETIC FORCES (Cont’d) A total force on a charge is given by Lorentz force equation: F qE u B The force is related to acceleration by the equation from introductory physics, F ma 13 MAGNETIC FORCES (Cont’d) To find a force on a current element, consider a line conducting current in the presence of magnetic field with differential segment dQ of charge moving with velocity u: dF dQu B But, dL u dt 14 MAGNETIC FORCES (Cont’d) So, dQ dF dL B dt Since dQ the line, dt corresponds to the current I in dF IdL B We can find the force from a collection of current elements F12 I 2dL2 B1 15 MAGNETIC FORCES (Cont’d) Consider a line of current in +az direction on the z axis. For current element a, IdL a Idz aa z But, the field cannot exert magnetic force on the element producing it. From field of second element b, the cross product will be zero since IdL and aR in same direction. Figure 3-28 (p. 128) 16 (a) Differential current elements on a line. (b) A pair EXAMPLE 7 If there is a field from a second line of current parallel to the first, what will be the total force? Figure 3-28 (p. 128) (a) Differential current elements on a line. (b) A pair of current-carrying lines will exert magnetic force on each other. 17 SOLUTION TO EXAMPLE 7 The force from the magnetic field of line 1 acting on a differential section of line 2 is: dF12 I 2 dL 2 B1 Where, 0 I1 B1 a 2 By inspection from figure, y, a a x Why?!?! 18 SOLUTION TO EXAMPLE 7 Consider dL 2 dza z , then: 0 I1 0 I1I 2 a x dF12 I 2dza z dz a y 2y 2y 0 0 I1I 2 F12 a y dz 2y L 0 I1I 2 L F12 ay 2y 19 MAGNETIC FORCES (Cont’d) Generally, 0 dL 2 dL1 a12 F12 I 2 I1 4 R12 2 • Ampere’s law of force between a pair of current- carrying circuits. • General case is applicable for two lines that are not parallel, or not straight. • It is easier to find magnetic field B1 by Biot-Savart’s law, then use F12 I 2dL2 B1 to find F12 . 20 EXAMPLE 8 The magnetic flux density in a region of free space is given by B = −3x ax + 5y ay − 2z az T. Find the total force on the rectangular loop shown which lies in the plane z = 0 and is bounded by x = 1, x = 3, y = 2, and y = 5, all dimensions in cm. Try to sketch this! 21 SOLUTION TO EXAMPLE 8 The figure is as shown. 22 SOLUTION TO EXAMPLE 8 (Cont’d) First, note that in the plane z = 0, the z component of the given field is zero, so will not contribute to the force. We use: F loop IdL x B Which in our case becomes with, I 30 A and B 3xa x 5 ya y 2 za z 23 SOLUTION TO EXAMPLE 8 (Cont’d) So, F 0.03 30 dxa x 3 xa x 5 y y 0.02 a y 0.01 0.05 30dya y 3x x 0.03 a x 5 ya y 0.02 0.01 30 dxa x x 3 xa x 5 y y 0.05 a y 0.03 0.02 30dya y x 3x x 0.01 a x 5 ya y 0.05 24 SOLUTION TO EXAMPLE 8 (Cont’d) Simplifying these becomes: F 0.03 0.05 0.01 0.01 0.02 0.02 0.03 0.05 30(5)(0.02)a z dx 30(3)(0.03) a z dy 30(5)(0.05)a z dx 30(3)(0.01) a z dy 0.06 0.081 0.150 0.027 a z N F 36a z mN 25