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Week 7 Lorentz Force Amperes Law Faradays Law. the Lorentz force is the force on a point charge due to electromagnetic fields. It is given in terms of.

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Presentation on theme: "Week 7 Lorentz Force Amperes Law Faradays Law. the Lorentz force is the force on a point charge due to electromagnetic fields. It is given in terms of."— Presentation transcript:

1 Week 7 Lorentz Force Amperes Law Faradays Law

2 the Lorentz force is the force on a point charge due to electromagnetic fields. It is given in terms of the electric field E and magnetic flux density B: F = q( E + v x B)

3 If a charged particle moves into a magnetic field, the particle will take on a curved trajectory: Notice that the force and the magnetic field are perpendicular to each other. This means that the B field does not do work on the charged particles or current.

4 The speed is just right, the forces will cancel each other out and the charge will move in a straight line

5 The radius of the trajectory is proportional to the mass-to-charge ratio This allows us to separate heavier ions from lighter ones

6 Since the force constantly changes the direction of the electron, the electron will start moving in a circular pattern preserving its initial speed v o. To derive an expression for the radius of rotation r in terms of B o, set the magnitude of F equal to the centrifugal force ½ m e r 2

7 An electron with constant velocity v = v o x enters in a magnetic field B = B o z. Calculate the initial magnetic force F exerted on the electron.

8 In semiconductors, the current carriers can be either electrons or electron holes. Electron holes (or simply holes) have a positive charge.

9 The forces on the charge carriers in a conductor in the presence of a magnetic field give rise to a voltage (Vab) across the width of the conductor.

10 Determine which terminal of the galvanometer is positive if the material is p type.

11 This law relates the magnetic flux density B to its source, the current I.

12 The line integral of the magnetic flux density B over a closed contour is proportional to the net current through the surface enclosed by the contour: Notice that the double integral is evaluated over the surface S enclosed by the closed curve C. The line integral is evaluated around the closed curve C.

13 The equation is correct in the special case where the electric field is constant (i.e. unchanging) in time. Otherwise, the equation must be modified.

14 The direction of the line differential and the direction of the surface differential are resolved using the right-hand rule: When the index-finger of the right-hand points along the direction of line integration, the outstretched thumb points in the direction that must be chosen for the vector area da, and current passing in that same direction must be counted as positive.

15 Using Amperes law, find B around a straight wire carrying a current I. Assume the wire is aligned with the z-axis.

16 The magnetic flux density B φ in a cylindrical region 0 r a carrying a current I z is given by B φ = μ o I z r 2π a 2 Determine the surface current density J z.

17 A cable consisting of an inner conductor, surrounded by a tubular insulating layer typically made from a flexible material with a high dielectric constant, all of which is then surrounded by another conductive layer (typically of fine woven wire for flexibility, or of a thin metallic foil), and then finally covered again with a thin insulating layer on the outside. The term coaxial comes from the inner conductor and the outer shield sharing the same geometric axis.

18 Coaxial cables are often used as a transmission line or radio frequency signals. In a coaxial cable the electromagnetic field carrying the signal exists only in the space between the inner and outer conductors. A coaxial cable provides protection of signals from external electromagnetic interference, and effectively guides signals with low emission along the length of the cable.

19 The curl of the magnetic flux density B is proportional to the current density that creates it: Again, this equation only applies in the case where the electric field is constant in time.

20 The absence of B fields around a coaxial cable results in no interference in nearby electrical equipment and wires. Show that if the current is the same magnitude in each direction, the magnetic field B outside the coaxial cable is zero. Find B for a { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/7/1632521/slides/slide_20.jpg", "name": "The absence of B fields around a coaxial cable results in no interference in nearby electrical equipment and wires.", "description": "Show that if the current is the same magnitude in each direction, the magnetic field B outside the coaxial cable is zero. Find B for a

21 Taking a rectangular path about which to evaluate Ampere's Law such that the length of the side parallel to the solenoid field is L gives a contribution (BL) inside the coil. The field is essentially perpendicular to the sides of the path, giving negligible contribution.

22 Show that in this idealized case, Ampere's Law gives B = μ o nI where n is the number of turns N per length L.

23 All of the loops of wire which make up a toroid contribute to the magnetic field in the same direction inside the toroid. The sense of the magnetic field is that given by the right-hand rule.

24 A more detailed visualization of the field of each loop can be obtained by examining the field of a single current loop. The current enclosed by the dashed line is just the number of loops times the current in each loop.

25 Show that the magnetic flux density B is given by B φ = μ o NI 2π r

26 Determine the expression for B due to an infinite plane sheet with uniform current density J. Assume that the sheet is on the x- y plane.

27 What if the electric field is not constant in time? Then we need to consider a displacement current.

28 The force exerted on the rod is given by F = I L x B

29 A straight wire carries a current I = 1 mA in the –x direction. The wire feels a force of 1 N per meter in the –z direction. Calculate the magnetic flux density B y.

30 Parallel conductors carrying currents in same direction will attract each other. If the currents are in opposite directions the conductors will repel each other.

31 The current I' is moving in the B field caused by the current I, so it experiences a force F = I' L x B.

32 Show that the magnetic force per unit length that one wire exerts upon the other is where

33 The ampere is defined to be the constant current which will produce an attractive force of 2 × 10 –7 Newtons per meter of length between two straight, parallel conductors of infinite length and negligible circular cross section placed one meter apart in free space.

34 Read Sections 5-1, 5-4, and 5-8 in your book. Solve end-of-chapter problems 5-12, 5-15, 5-16, 5.20, 5.21.


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