Ch 34 Electromagentic Fields and Waves Reading Quiz

Ch – 34 Electromagentic Fields and Waves

Reading Quiz – Ch. 34 • • • 1. Experimenter A creates a magnetic field in the laboratory. Experimenter B moves relative to A. Experimenter B sees: a. the same magnetic field. b. a magnetic field of different strength. d. an electric field. e. both a magnetic and an electric field. 2. Which of the following was not discussed in the sections 1 -4 of this chapter: a. Gauss’s Law b. The Lorentz Force c. Electrical Conduction in Gases d. Galilean Transformations

Learning Objectives – Ch 34 • To understand that electric and magnetic fields are interdependent. There’s just a single electromagnetic field that presents different faces, in terms of and to different observers. • Electromagnetic fields obey four general laws, called Maxwell’s equations. • Electromagnetic fields can exist without source charges or currents in the form of a selfsustaining electromagnetic wave. • Maxwell’s equations predict that all electromagnetic waves travel at the same speed.

Electromagnetic fields away from q direction given by right hand rule Field vectors for E and B

Field Lines

The Lorentz Force A charge moves through a region where there is both an external electric and an external magnetic field.

The Lorentz Force The net force on the charge is due to both these forces. This is known as the Lorentz Force law:

Case 1: Moving or Stationary? • Bill says there will be a B field generated by the moving charge, in addition to the E field generated by the charge • But Sharon claims there will be an E field, but not a B field, because she “sees” the charge as being stationary relative to her*. *This story is a bit foolish until you imagine Sharon being on the Moon, or at least in a space station, so she can really believe the charge in her hand is stationary.

Case 2: What causes the force? • Bill turns on a constant, external magnetic field. Bill claims there should be a upward force due to B for the moving charge. • Sharon feels the force on the charge, but claims it must be from something else because stationary charges can’t experience a force in a constant B field.

Galilean Relativity – Review of ch. 6 -4 • Reference frame S’ (think of a space station) moves with constant velocity V relative to reference frame S (think of Earth). • An observer at rest in S sees the origin of S’ whiz by at V. Of course the guy in the S’ sees S go by at –V.

Galilean Relativity • Observers in both reference frames see a particle whiz by. • Observer in S would say its velocity is v. Observers in S’ would say the velocity is v’ where: • v’ = v-V and • v = v’ + V

Galilean Relativity v’ = v-V and v=v+V How does the particle’s acceleration compare in the two frames? dv’/dt = dv/dt – d. V/dt since V is constant: a’ = a and F’ = F Observers in all inertial reference frames agree about the force. What differs is how they perceive the B and E fields that cause the force.

Case 1: Moving or Stationary? • Bill measures the E field and the B field due to the moving charge in reference frame S. • Shouldn’t Sharon, in frame S’, measure the same fields? After all, aren’t those fields just “there”?

Let’s experiment • Establish a region where Eext is known to be 0, but there is a non-zero Bext • Send a moving charge through B. Since there is no Eext

Case 2: B causes the force for an observer in frame S • Bill can measure an upward force and from the magnetic field: and v = V:

How about Sharon, in S’? • She observes the same magnitude upward force, one that is proportional to q. She concludes that, whethere is a Bext or not, there must be an Eext, because from her point of view, the charge is at rest.

How about Sharon, in S’? • And since there is no v’ = 0….

Case 2: E’ causes the force for an observer in frame S’ • Sharon can measure an upward force and from the external E field, even though Bill swears there is none.

How do the fields relate • But both observers agree on the magnitude of the force, therefore: • E’ = V X B

The General Case • An experimenter in S creates external fields, B and E. • A charge moving with velocity v relative to S experiences a Lorentz force of:

The General Case • The charge is at rest in frame S’ so the force in S’ can be due only to the E field: • Since both observers agree about F: • E’ = E + V x B

Transformation for E’ E’ = E + V x B This expression transforms electric and magnetic fields in S into the electric field measured in S’.

Transformation for B’ The charge at rest in frame S produces an E field but no B field: B=0 What are the fields measured by Sharon in frame S’? She sees a charge moving with velocity v’ = -V.

Transformation for B’ Since B = 0: E’ = But Sharon also measures a B’ field since in her frame the charge is moving: B’ = - (u 0/4π)q. V/r 2 X

Transformation for B’ Since B = 0: E’ = But Sharon also measures a B’ field since in her frame the charge is moving: B’ = - (u 0/4π)q. V/r 2 X = -ε 0μ 0 V X

Transformation for B’ Since B = 0: E’ = But Sharon also measures a B’ field since in her frame the charge is moving: B’ = - (u 0/4π)q. V/r 2 X = -ε 0μ 0 V X

Transformation for B’ B’ = -ε 0μ 0 V X E. If observers in frame S create a magnetic field B in addition to E field, then: B’ = B - ε 0μ 0 V X E.

Galilean Field Transformations

Which diagram shows the fields in frame S’

Answer is b

A conductive loop in frame S In frame S, a conductive loop is moving with velocity v, into a “stationary” magnetic field. The B field causes an force on the moving charges. The resulting is charge separation and an emf. Note there is no pre -existing electric field. ε = v. LB E = dε/d. L = v. B

Same loop in frame S’ In frame S’, the loop is stationary, but B travels to the left with velocity -v. An observer in this frame agrees that there is a force on the charge, but since the loop is stationary, the force must be due to an electric field, E’.

The same loop in frame S’ Using the field transformations (recall E in frame S equals 0):

The same loop in frame S’ The observer in frame S’ sees the same B field as the observer in S, but in addition he sees an E field as well. The B field does not exert a force in S’.

The same loop in frame S’ The electric force, E’ in frame S’ exerts the same magnitude force and generates the same emf as the B field in frame S.

E’ is the induced electric field of Faraday’s Law In frame S, this electric field is the result of the charge separation caused by the magnetic force. In frame S’ it is the cause: E = v. B

E’ is a non-Coulomb Field It is not created by static electric charges, but rather by a changing magnetic field.

Calculating the emf Recall W = -∆U and ∆V = ∆U/q ε = W/q This is another example of a line integral. Two cases are of interest.

Calculating the emf For a closed curve of length, L: 1. If E is everywhere perpendicular to the integration path, ε = 0. 2. If E is everywhere tangent to the integration path, ε = EL.

Sign Convention Use right hand rule with thumb showing positive direction of flux and fingers showing positive direction for emf (current direction)

Faraday’s Law, revisited Using this convention, ε and dΦ/dt will always have opposite signs. Therefore: ε = - dΦm/dt A full statement of Faraday’s Law:

Displacement Current Ampere’s law relates the integral of B around curve C to the current passing through surface S. This is for the case of a flat surface.

Displacement Current It still holds if the surface is not flat. In both cases shown, both S 1 and S 2 are bounded by curve C. In both cases, Ithrough-1 = Ithrough-2

Displacement Current • If we use this logic for the plates of a capacitor, it fails us since Ithrough-1 does not equal Ithrough-2. • There is no current through S 2 but there is a changing electric flux through S 2 as capacitor charges

Displacement Current (Ithrough S 1) • • • Φe = EA for capacitor E = Q/(ε 0 A) Φe = Q/ε 0, I = d. Q/dt d Φe /dt = I / ε 0 d Φe /dt is called the displacement current idisp

Ampere-Maxwell Law

Induced B and E fields • “Ordinary” electric fields are created by charge but we can create an electric field by having a changing magnetic field. • “Ordinary” magnetic fields are created with a current, but we can create a magnetic field by having a changing electric field.

Faraday’s Law Problem(#12) • The graph shows the current as a function of time for a 20 -cm long, 4. 0 cm diameter solenoid with 400 turns Draw a graph of the induced E field as a function of time at a point 1. 0 cm from the axis of the solenoid.

Faraday’s Law Problem(#12) • For 0<t<0. 1 s, E = -6. 28 x 10 -4 V/m • For 0. 1<t<0. 2 s, E = +3. 14 x 10 -4 V/m

Displacement Currents (#18) • A 10 -cm diameter plate capacitor has a 1. 0 mm spacing. The E field increases at the rate of 1. 0 x 106 V/m-s. What is the magnetic field strength: • A: on the axis • B: 3. 0 cm from the axis • C. 7. 0 cm from the axis

Displacement Currents (#18) • A 10 -cm diameter plate capacitor has a 1. 0 mm spacing. The E field increases at the rate of 1. 0 x 106 V/m-s. What is the magnetic field strength: • A: 0 T • B: 1. 67 x 10 -13 T • C: 1. 98 x 10 -13 T

Sign Convention Recall Faraday’s Law: ε = |dΦm/dt|, direction given by Lenz’ Law. Use of Lenz’ Law means we can’t use Cartesian conventions of positive and negative to establish sign of Φm, or ε. So make an arbitrary choice for Φ and a convention for ε:

Sign Convention • • Curl right hand around surface in most painfree direction. The direction in which your thumb points is positive. The area vector A, always points in the positive direction. Φ is the vector dot product of B and A; it will be positive or negative, depending on the direction of B.

Sign Convention • • Don’t forget that its dΦm/dt that is important, so adjust sign according to whether Φm increases or decreases with time. A positive emf creates an induced current in the direction of your fingers; a negative emf creates a current in the opposite direction.