1. 1 Expression for curl by applying Ampere’s Circuital Law might be too lengthy to derive, but it can be described as: The expression is also called the point form of Ampere’s Circuital Law, since it occurs at some particular point. AMPERE’S CIRCUITAL LAW (Cont’d)

2. 2 The Ampere’s Circuital Law can be rewritten in terms of a current density, as: Use the point form of Ampere’s Circuital Law to replace J, yielding: This is known as Stoke’s Theorem. AMPERE’S CIRCUITAL LAW (Cont’d)

3. 3 3.3MAGNETIC 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: Where μ is the permeability with:

4. 4 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:

5. 5 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 MAGNETIC FLUX DENSITY (Cont’d)

6. 6 The magnet cannot be divided in two parts, but it results in two magnets. The magnetic pole cannot be isolated.

7. 7 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 : By applying divergence theorem, the point form of Gauss’s Law for static magnetic fields:

8. 8 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 a z direction. Try to sketch this!

9. 9 SOLUTION TO EXAMPLE 6 The relation between B and H is: To find flux crossing the portion, we need to use: where d S is in the a φ direction.

10. 10 So, Therefore, SOLUTION TO EXAMPLE 6 (Cont’d)

11. 11 3.4MAGNETIC FORCES Upon application of a magnetic field, the wire is deflected in a direction normal to both the field and the direction of current.

12. 12 MAGNETIC FORCES (Cont’d) The force is actually acting on the individual charges moving in the conductor, given by: By the definition of electric field intensity, the electric force F e acting on a charge q within an electric field is:

13. 13 A total force on a charge is given by Lorentz force equation : MAGNETIC FORCES (Cont’d) The force is related to acceleration by the equation from introductory physics,

14. 14 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 : But,

15. 15 So, Since corresponds to the current I in the line, MAGNETIC FORCES (Cont’d) We can find the force from a collection of current elements

16. 16 Consider a line of current in + a z direction on the z axis. For current element a, 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 Id L and a R in same direction. MAGNETIC FORCES (Cont’d)

17. 17 EXAMPLE 7 If there is a field from a second line of current parallel to the first, what will be the total force?

18. 18 The force from the magnetic field of line 1 acting on a differential section of line 2 is: Where, By inspection from figure, Why?!?! SOLUTION TO EXAMPLE 7

19. 19 Consider, then: SOLUTION TO EXAMPLE 7

20. 20 Generally, 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 B 1 by Biot-Savart’s law, then use to find F 12. MAGNETIC FORCES (Cont’d)

21. 21 EXAMPLE 8 The magnetic flux density in a region of free space is given by B = −3 x a x + 5 y a y − 2 z a z 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!

22. 22 The figure is as shown. SOLUTION TO EXAMPLE 8

23. 23 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: Which in our case becomes with, and

24. 24 So, SOLUTION TO EXAMPLE 8 (Cont’d)

25. 25 Simplifying these becomes: SOLUTION TO EXAMPLE 8 (Cont’d)