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 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 3.3MAGNETIC 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 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 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 The magnet cannot be divided in two parts, but it results in two magnets. The magnetic pole cannot be isolated.

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

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

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 Consider, then: SOLUTION TO EXAMPLE 7

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 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 The figure is as shown. SOLUTION TO EXAMPLE 8

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 So, SOLUTION TO EXAMPLE 8 (Cont’d)

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