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## Monday, August 31, 2009

### AP Physics C – Multiple Choice Questions (for Practice) on Maxwell’s Equations and Electromagnetic Waves

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As promised in the post dated 26th August 2009, I give below a few typical multiple choice practice questions for AP Physics C students:
(1) The displacement current in the dielectric between the plates of a 0.1 μF capacitor is 1 mA. The potential difference between the plates of this capacitor must be changing at the rate of
(a) 104 volt/second
(b) 103 volt/second
(c) 102 volt/second
(d) 10 volt/second
(e) 1 volt/second
This is a simple question contrary to the impression of some of you at the first glance! Whether it is conduction current or displacement current, we have the usual expression for electric current (i) as
i = dQ/dt
Since the charge Q = CV, we have i = CdV/dt
This gives dV/dt = i/C = 1 mA /0.1 μF = 10–3 ampere /10–7 farad = 104 volt/second [Option (a)].
(2) A 10 μF electrolytic capacitor connected to a 6 V power supply is fully charged. The displacement current flowing through the dielectric is
(a) 100 mA
(b) 10 mA
(c) 1mA
(d) 0.1 mA
(e) zero
The correct option is (e) since there is no charging current in the case of the fully charged capacitor.
(3) The electric and magnetic fields in a radio wave propagating along the z-direction are given respectively by
Ex= E0 sin (kz–ωt ) and
By= B0 sin (kz–ωt )
where E0 and B0 are the amplitudes, k is the magnitude of the wave vector (propagation vector) and ω is the angular frequency. How is E0 related to B0?
(a) E0 = B0 kω
(b) E0 = B0k/ω
(c) E0 = B0√(kω)
(d) E0 = B0
(e) E0 = B0ω/k
The electric and magnetic fields in the wave are varying simple harmonically and the speed of the wave is given by c = ω/k. [Note that the speed of a wave propagating in the z-direction is given by the ratio of coefficient of t to the coefficient of z. [See the post dated 9th April 2009 ‘AP Physics B- Wave Motion (including sound)- Points to be Noted’ which you can access by clicking on the label ‘waves’ below this post.
Since E = Bc we have E0 = B0c = B0ω/k
(4) The charge on the plates of a capacitor of plate area 0.1 m2 is 0.05 coulomb and is decreasing at the rate of 0.5 coulomb per second at an instant t during its discharge. What is the displacement current at the instant t?
(a) 0.005 A
(b) 0.05 A
(c) 0.5 A
(d) 5 A
(e) Data insufficient for arriving at an answer
Data is more than sufficient. The plate area given in the question is not required and it serves the purpose of a distraction. The displacement current is the time rate of change of charge (dQ/dt) on the plates and is equal to 0.5 ampere [Option (c)].
[If the capacitor is an ideal parallel plate capacitor and you are required to find the magnitude of the displacement current density, the answer is 0.5 A/0.1 m2 = 5 Am–2]
(5) At a particular point in space and time, the magnetic field in a plane electromagnetic wave traveling in free space along the positive x-direction is given by B = 3.1×10–8 k T (where i, j, k are unit vectors along the x, y, z directions as usual). The electric field at the same point at the same instant is
(a) 1.033 j Vm–1
(b) 9.3 i Vm–1
(c) – 9.3 j Vm–1
(d) 1.033 k Vm–1
(e) 9.3 j Vm–1
The magnitude of the electric field is given by E = Bc = 3.1×10–8 ×3×108 Vm–1 = 9.3 Vm–1. Since the wave is proceeding along the positive x-direction and the magnetic field vector is along the positive z-direction (as indicated by the unit vector k), the electric field must be along the positive y-direction. [Note that the cross product vector E×B should point along the direction of propagation].
Therefore, E = 9.3 j Vm–1.
(6) A microwave signal traveling through the earth’s atmosphere has an electric field of amplitude 0.02 Vm–1. The intensity (power flow per unit area) of this signal is (Permittivity of free space, ε0 = 8.85×10–12 C2 N–1 m–2)
(a) 5.31×10–2 Wm–2
(b) 5.31×10–3 Wm–2
(c) 5.31×10–6 Wm–2
(d) 5.31×10–7 Wm–2
(e) 5.31×10–8 Wm–2
The intensity (I) which is the power flow through unit area (with the plane of the area perpendicular to the direction of propagation) is given by
I = ε0 c Em2/2 where c is the speed of the signal and Em is the amplitude of the electric field.
Therefore, I =(8.85×10–12 ×3×108×0.0004) / 2 = 5.31×10–7 Wm–2

## Wednesday, August 26, 2009

### AP Physics C - Maxwell’s Equations and Electromagnetic Waves

This post is meant for AP Physics C aspirants who are required to have some idea about Maxwell’s equations and their consequence.

Maxwell’s equations are basically the mathematical statements of

(i) Gauss’s law in electricity

(ii) Gauss’s law in magnetism

(iii) Faraday’s law of electromagnetic induction and

(iv) Ampere-Maxwell law.

The last one is the well known Ampere’s law with Maxwell’s modification for incorporating displacement current (which can flow even through empty space), in addition to conduction current (which flow through conductors). Here are Maxwell’s equations:

1. ∫closed surfaceE.dA = Q/ε0

[This is Gauss’s law in electricity which states that the flux of the electric field E through any closed surface, that is, the surface integral of E.dA over any closed surface is 1/ε0 times the total charge Q enclosed by the surface. Note that E is the electric field vector present at an elemental area vector dA of the closed surface].

2.closed surfaceB.dA = 0

[This is Gauss’s law in magnetism which states that the magnetic flux through a closed surface is zero].

3. closed pathE.d = – dФB/dt

[This is Faraday’s law of electromagnetic induction which states that the induced emf in a circuit is equal to the time rate of change of magnetic flux ФB. The negative sign is because of Lenz’s law which states that the induced emf opposes the change of the magnetic flux].

4. closed pathB.d = μ0ic + μ0ε0dФE/dt

[This is Ampere’s circuital law with Maxwell’s modification. The firest term, μ0ic on the right hand side contains the conduction current ic. The second term, μ0ε0(dФE/dt) was added by Maxwell to incorporate the displacement current id = ε0(dФE/dt). Note that the displacement current is produced because of the time rate of change of the electric field].

Maxwell’s equations given above are in the integral form. The differential form of Maxwell’s equations can be easily obtained by applying Gauss’s divergence theorem and Stokes theorem. Thus we have the following equations:

(i) ∫closed surfaceE.dA = Q/ε0 becomes v div Edv = (1/ε0)v ρdv, on applying Gauss divergence theorem to the left hand side of the equation and by putting Q = v ρdv where ρ is the volume charge density. The volume integration is done over the volume ‘v’ enclosed by the closed surface. Therefore,

div E = ρ/ε0

This can be written also as

div D = ρ

where D is the electric displacement vector which in free space is given by D = ε0E.

(ii) The second equation, closed surfaceB.dA = 0 similarly becomes

div B = 0

(iii) The 3rd equation, closed path E.d = – dФB/dt becomes s curl E.dA = ∫s (d/dt) B. dA, on applying Stokes theorem to the left hand side of the equation and by putting the magnetic flux ФB = s B. dA. The surface integration is performed over the entire area A enclosed by the closed path. Therefore,

curl E = – dB/dt

(iv) The 4th equation, ∫closed pathB.d = μ0ic + μ0ε0 (dФE/dt) becomes s curl B.dA = ∫s μ0J. dA + ∫s μ0ε0(d/dt) E. dA, on applying Stokes theorem to the left hand side of the equation and by putting ic = ∫s J . dA and ФE= s E. dA. Therefore,

curl B = μ0J + μ0ε0(dE/dt) = μ0 (J + dD/dt) since D = ε0E for free space.

Displacement current:

The concept of the displacement current was introduced by Maxwell from his understanding that all electric currents must be closed. For instance, in the charging of a capacitor, a conduction current ic flows in the wires connecting the capacitor to the charging battery and an equal (total) displacement current flows through the dielectric (or free space) in between the capacitor plates. The conduction current in the connecting wire and the displacement current in the space between the plates of the capacitor make a closed current circuit.

The displacement current (i) as well as the conduction current is given, as usual, by

i = dQ/dt

But Q = CV where C is the capacitance and V is the voltage across the capacitor. Therefore dQ = CdV so that the displacement current i = CdV/dt.

In the simple case of a parallel plate capacitor with air or free space between the plates, C = ε0A/d where A is the area of the plates and d is the separation between the plates. Further, V = Ed where E is the electric field between the plates. Therefore, displacement current i = (ε0A/dd×dE/dt = ε0A dE/dt = ε0dФE/dt, on substituting for the electric flux ФE = AE.

The above steps show that the quantity ε0dФE/dt indeed represents the displacement current.

Electromagnetic Waves:

Electromagnetic waves are produced by accelerated charges. An oscillating electric charge produces an oscillating electric field in space, which produces an oscillating magnetic field. But an oscillating magnetic field is a source of oscillating electric field. Therefore an oscillating electric charge can produce oscillating electric and magnetic fields which regenerate each other and an electromagnetic wave propagates through the space.

The electric field and the magnetic field in an electromagnetic wave are perpendicular to each other. (Note that in charging a capacitor, the electric field in the space between the capacitor plates is directed perpendicular to the plates where as the magnetic field produced by the displacement current flowing through the space between the capacitor plates is along circles around the electric field lines).

In the case of a plane electromagnetic wave propagating along the positive z-direction, the electric field is along the positive x-direction and the magnetic field is along the positive y-direction.

Remember:

(i) The vectors E and B in an electromagnetic wave are perpendicular to each other and so oriented that the vector product (cross product) E×B points in the direction of propagation of the wave.

As an example of the application of this rule, suppose the electric field vector is along the negative z-direction and the magnetic field vector is along the positive x-direction. This wave has to be propagating along the negative y-direction (Fig.).

(ii) The magnitudes (E and B) of the electric field and the magnetic field in an electromagnetic wave are related as

B = E/c

where c is the speed of electromagnetic waves.

(iii) The speed (c) of electromagnetic waves in free space is given by

c = 1/√(μ0 ε0)

(iii) The speed (v) of electromagnetic waves in a medium of permittivity ε and permeability μ is given by

v = 1/√(μ ε)

Since μ = μ0 μr and ε = ε0 εr where μr and εr are respectively the relative permeability and the relative permittivity of the medium, we have

v = c/√(μr εr)

(iv) The energy density (energy per unit volume) in the region of space through which an electromagnetic wave propagates is due to the electric and magnetic fields associated with the wave. The energy density (UE) due to the electric field is given by

UE = ½ ε0Erms2 where Erms is the root mean square value of the electric field. [We use the root mean square value since the field is oscillating].

Similarly the energy density (UB) due to the magnetic field is given by

UB = ½ (Brms2/μ0) where Brms is the root mean square value of the magnetic field.

UE and UB are equal since Brms = Erms/c = Erms√(μ0 ε0)

The energy density (U) due to the electromagnetic wave is the sum of the above energy densities and is given by

U = UE + UB = ½ ε0Erms2 + ½ (Brms2/μ0)

Since UE = UB, we have

U = ε0Erms2 = Brms2/μ0

In the case sinusoidally oscillating electric and magnetic fields Erms = Em/√2 and Brms = Bm/√2 so that the energy density can be written as

U = ε0Em2/2 = Bm2/2μ0

(v) The intensity (I) of the electromagnetic wave, which is the power flow through unit area, that is, the energy flowing per second through unit area (the plane of the area held perpendicular to the direction of propagation of the wave) is given by

I = Uc = ε0c Em2/2 = (Em2/2)(ε0/μ0) since c =1/√(μ0 ε0)

This can also be written as

I = Em2/2cμ0

(vi) The power flow through unit area is described by Poynting vector S given by

S = E×B/μ0

Poynting vector S is directed along the direction of propagation of the wave. Since E and B are perpendicular to each other, Poynting vector has magnitude EB/μ0 which is equal to E2/cμ0. Note that this quantity represents the instantaneous power flow through unit area.

In the next post we will discuss questions in this section. Meanwhile see these posts at physicsplus.