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3. 1.Khurram Shahzad 2.DanIs H ussain 3.Bukhtyar Ali 4.Shah Mehmood 5.Farrukh Ali 6.Usman Akhtar

4. www.BZUpages.com KHURRAM SHAHZAD BS(IT)3rd ROLL# 07-32 Ampere’s Law Presented To:- Dr. Tariq Bhatti

5. www.BZUpages.com Ampere’s Law Andre Marie Ampere (1775-1836)

6. www.BZUpages.com “ Ampere's Law states that for any closed loop path, the sum of the quantities (B.ds) for all path elements into which the complete loop has been divided is equal to the product of µ 0 and the total current enclosed by the loop.” DEFINITION “ Relationship between magnetic field and electric current.” OR

7. www.BZUpages.com EXPERIMENT

8. Take a circle of radius r as the Ampere Loop.

9. www.BZUpages.com In the Figure:  Dot(.) represents a wire which is placed perpendicular to the plane in the form of circle.  Inner circle is the closed path or Amperian path.  “r” is the radius of closed path.  “ds” is the displacement across the closed Path.

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11. Direction of magnetic field can be shown by drawing a tangent at any point.

12. www.BZUpages.com Thus the dot product of B & the short vector ds is: So where Here -------- (A)

13. www.BZUpages.com By Biot–Savart law So as By substituting the value of B in equation (1) we get: ------ (1)

14. www.BZUpages.com So equation (2) can be written as Here …… (2) By equation (2),we can say that magnetic field depends upon current and permeability of free space.

15. www.BZUpages.com DanIs H ussain BS( IT ) 3rd ROLL # 0 7- 0 9 Presented To:- Dr. Tariq Bhatti Application of Ampere’s Law

16. www.BZUpages.com  A long straight current carrying wire is placed In a uniform magnetic field.  “a” is the radius of wire.  “R” is the radius of circle.  “R” > “a”.  “B” is constant at each point on circular path. Experiment :

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18. As magnitude of “B” has the same value at each point on the circular path. So equation A becomes: Here as So ------ (3)

19. www.BZUpages.com …… (5) By substituting the value of B in equation (3) we get: By Comparing equation (3) & equation (4) we get: (R>a) This is magnetic field outside the wire. …… (4)

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21. Now we calculate the effect of magnetic Field Inside the surface of wire. By equation (5): Here J=current/cross section area of closed path= So equation (7) becomes …… (6) …… (7)

22. www.BZUpages.com By substituting the value of in equation (6) we get: (R<a) This is the magnitude of magnetic field due to the current inside the surface of wire. Here BάI BάR Bά1/a 2

23. www.BZUpages.com BUKHTYAR ALI BS(IT)3rd ROLL# 07-18 Presented To:- Dr. Tariq Bhatti The Displacement Current & Ampere's law

24. www.BZUpages.com Displacement Current (a)An Amperian loop encloses a surface through which passes a wire carrying a current (b) The same Amperian loop encloses a surface that passes between the capacitor plates. No conduction current passes through the surface.

25. www.BZUpages.com Circular metal plates

26. www.BZUpages.com  Let ‘s1’ and ‘s2’ be two closed surface. Figure shows a cross-section of the capacitor and the electric field in the region Between plates. Here flux is Ø E =EA Since the field exists only in the region between plates.

27. www.BZUpages.com E=Charge density/permeability of free space ----------------- (2) E= σ / Є 0 σ= Q/A E = Q/AЄ 0 By substituting the value of σ in equation (2) we get: i.e. Q = Є 0 EA ----------------- (1)

28. www.BZUpages.com Electric Flux=Ø E =EA As we know that So Q=Є 0 Ø E By substituting the value of Q in equation (1) we get: as so Current displacement

29. www.BZUpages.com By Ampere law By adding displacement current in Ampere law By substituting the value of displacement current we get: -------- (3)

30. www.BZUpages.com Equation (3) is the modified form of Ampere's Law.  The displacement current and the modified form of Ampere's Law is an essential part in study of Electromagnetic waves.  The effect of the displacement current is negligible in circuits with slowly varying Currents and fields except the following described example.

31. www.BZUpages.com Maxwell’s Equations The four fundamental equations of electromagnetism, called Maxwell’s equation. NameEquation Gauss’ law for electricity Gauss’ law for magnetism Faraday’s law Ampare-Maxwell law These equations are the basis of many of the equations we see in “charge particle” to “optics”.

32. www.BZUpages.com Shah Mahmood BS(IT)3rd ROLL# 07- 41 Presented To:- Dr. Tariq Bhatti Force between Current

33. www.BZUpages.com Force between Current Suppose we have a two current carrying wire placed in the magnetic field. We can determine the magnetic force that one current carrying wire exerts on other. Consider the arrangement of two parallel wires I (one) & I (two) separated by a distance R. The magnitude of the magnetic field is given by.

34. www.BZUpages.com EXPERIMENT

35. www.BZUpages.com Once the magnetic field has been calculated, the magnetic force expression can be used to calculate the force. The direction is obtained from the right hand rule. Note that two wires carrying current in the same direction attract each other, and they repel if the currents are opposite in direction.magnetic force expressionright hand rule Force between Current

36. www.BZUpages.com The calculation below applies only to long straight wires, but is at least useful for estimating forces in the ordinary circumstances of short wires. Once you have calculated the force on wire 2, of course the force on wire 1 must be exactly the same magnitude and in the opposite direction Force between Current

37. www.BZUpages.com The magnetic field of an infinitely long straight wire can be obtained by applying Ampere's law. Ampere's law takes the formmagnetic field straight wire Ampere's law and for a circular path centered on the wire, the magnetic field is everywhere parallel to the path. The summation then becomes just The constant μ 0 is the permeability of free space.permeability

38. www.BZUpages.com F=I 2 LB 1………………………( 1) B 1 =µ 0 I 1 F=µ 0 I 1 I 2 L /2πR Force exerted on wire 2 is By substituting the value of B in equation (1) we get:

39. FARRUKH ALI Roll no 07-44 Gauss’s LAW

40. www.BZUpages.com Flux Flux in Physics is used to two distinct ways. The first meaning is the rate of flow, such as the amount of water flowing in a river, i.e. volume per unit area per unit time. Or, for light, it is the amount of energy per unit area per unit time. Let’s look at the case for light:

41. www.BZUpages.com Area Vector Represent an area as a vector, of length equal to the area. The flux of light through a hole of area  A is proportional to the area, and the cosine of the angle between the light direction and this area vector.  If we use a vector to represent the light energy per unit time, then the light out of the hole is. In this case it is negative which means the light flux is into the hole.

42. www.BZUpages.com Flux of Electric Field Like the flow of water, or light energy, we can think of the electric field as flowing through a surface (although in this case nothing is actually moving). We represent the flux of electric field as  (greek letter phi), so the flux of the electric field through an element of area  A is When we have a complicated surface, we can divide it up into tiny elemental areas:

43. USMAN AKHTAR Roll no 07-17 Gauss’s LAW

44. www.BZUpages.com Gauss’ Law We are going to be most interested in closed surfaces, in which case the outward direction becomes self-evident. We can ask, what is the electric flux out of such a closed surface? Just integrate over the closed surface: The symbol has a little circle to indicate that the integral is over a closed surface. The closed surface is called a Gaussian surface, because such surfaces are used by Gauss’ Law, which states that: Gauss’ Law The flux of electric field through a closed surface is proportional to the charge enclosed. Flux positive => out Flux negative => in

45. www.BZUpages.com Mathematical Statement of Gauss’ Law The constant of proportionality in Gauss’ Law is our old friend  . Coulomb’s constant is written ? Coulomb’s constant is written ? We can see it now by integrating the electric flux of a point charge over a spherical gaussian surface. r q enc Solving for E gives Coulomb’s Law.

46. www.BZUpages.com Example of Gauss’ Law Consider a dipole with equal positive and negative charges. Imagine four surfaces S 1, S 2, S 3, S 4, as shown. S 1 encloses the positive charge. Note that the field is everywhere outward, so the flux is positive. S 2 encloses the negative charge. Note that the field is everywhere inward, so the flux through the surface is negative. S 3 encloses no charge. The flux through the surface is negative at the upper part, and positive at the lower part, but these cancel, and there is no net flux through the surface. S 4 encloses both charges. Again there is no net charge enclosed, so there is equal flux going out and coming in—no net flux through the surface.

47. www.BZUpages.com Field Inside a Conductor We can use Gauss’ Law to show that the inside of a conductor must have no net charge. Take an arbitrarily shaped conductor, and draw a Gaussian surface just inside. Physically, we expect that there is no electric field inside, since otherwise the charges would move to nullify it. Since E = 0 everywhere inside, E must be zero also on the Gaussian surface, hence there can be no net charge inside. Hence, all of the charge must be on the surface (as discussed in the previous slide). If we make a hole in the conductor, and surround the hole with a Gaussian surface, by the same argument there is no E field through this new surface, hence there is no net charge in the hole.

48. www.BZUpages.com Field Inside a Conductor We have the remarkable fact that if you try to deposit charge on the inside of the conductor... The charges all move to the outside and distribute themselves so that the electric field is everywhere normal to the surface. This is NOT obvious, but Gauss’ Law allows us to show this! There are two ideas here Electric field is zero inside conductors Because that is true, from Gauss’ Law, cavities in conductors have E = 0

49. www.BZUpages.com Use Gauss’ Law to Find Out Gaussian Surface Is E = 0 in the conductor? Yes, because as before, if there were an electric field in the conductor, the charges would move in response (NOT Gauss’ Law). If we enlarge the Gaussian surface so that it is inside the conductor, is there any net charge enclosed? It looks like there is, but there cannot be, because Gauss’ Law says E = 0 implies q enc = 0 ! How do we explain this? There must be an equal and opposite charge induced on the inner surface.

50. www.BZUpages.com Summary Electric flux is the amount of electric field passing through a closed surface. Flux is positive when electric field is outward, and negative when electric field is inward through the closed surface. Gauss’ Law states that the electric flux is proportional to the net charge enclosed by the surface, and the constant of proportionality is  . In symbols, it is

51. www.BZUpages.com http://www.rpi.edu/~persap/P2F07_ persans/software/gauss/examples.ht ml WWW.WIKIPEDIA.COM WWW.ASK.COM WWW.HYPERPHYSICS.COM

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