Kamal A.A. 1000 Solved Problems in Classical Physics: An Exercise Book

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630 13 Electromagnetism I

qE + qv

× B = 0

or E =−V

× B

E =−v

B (∵ v

⊥B)

E =

= v

B =

Wtne

∴ n =

etV

100 × 10

1.6 × 10

−19

× 10

−3

× 5 × 10

−6

= 1.25 ×10

13.85 If R

is the Hall coefﬁcient, σ the electrical conductivity, then the mobility

μ is given by

μ = R

σ = (−7.3 × 10

−5

)(2 × 10

) =−0.146 m

/V/s

The magnitude is 0.146 m

/V/s

Chapter 14

Electromagnetism II

Abstract Chapters 13 and 14 are devoted to electromagnetism concerned with

motion of charged particles in electric and magnetic ﬁelds, Lorentz force, cyclotron

and betatron, magnetic induction, magnetic energy and torque, magnetic dipole

moment, Faraday’s law, Hall effect, RLC circuits, resonance frequency, Maxwell’s

equations, electromagnetic waves, Poynting vector, phase velocity and group veloc-

ity, dispersion relations, waveguides and cut-off frequency.

14.1 Basic Concepts and Formulae

Self-Inductance (L)

L =

Nϕ

Total ﬂux linkage

Current linked

(14.1)

ξ =−L

(14.2)

For a long solenoid or toroid

L =

(14.3)

where N is the number of turns, A is the area of cross-section of each turn and l is

the length of the coil.

The energy stored in an inductor is

W =

(14.4)

L–R Circuit

Differential equation. L

+iR = ξ (for charging process) (14.5)

631

632 14 Electromagnetism II

Solution i =



1 − e

−



(14.6)

Inductive time constant

= L/R (14.7)

Differential equation L

+ iR = 0 (discharging process) (14.8)

Solution i =

−Rt /L

(14.9)

C–R Circuit

Ri + q/c = ξ(charging) (14.10)

q = q

(1 − e

−t/RC

) (14.11)

Capacitive Time Constant

τ = RC (14.12)

Ri + q/C = 0 (discharging) (14.13)

q = q

−t/RC

(14.14)

Inductors in Series

= L

+ L

+ 2M (coil currents in the same sense) (14.15)

= L

+ L

− 2M (coil currents in the opposite sense) (14.16)

where M is the mutual inductance

= L

+ L

(inductors well separated) (14.17)

Inductors in Parallel

= L

/(L

+ L

) (14.18)

M = (L

)

1/2

(inductors are closely placed) (14.19)

The Alternating Current (AC)

The effective or root-mean-square value:

= I

√

2 ( 14.20)

where I

is the peak current

14.1 Basic Concepts and Formulae 633

Reactance: Inductive X

= ωL, Capacitance X

= 1/ωC (14.21a)

Impedance Z =



+ (X

− X

)

(14.21b)

RLC Series Resonance Circuit

= I





ωL −

ωC



+ R

(14.22)

The phase α is given by the relation

tan α =

− X

ωL −

ωC

(14.23)

(i) α is positive if ωL > 1/ωC, and V leads I.

(ii) α is negative if ωL < 1/ωC, and V lags behind I.

(iii) α is zero if ωL = 1/ωC, and V is in phase with I .

2π

√

(resonance frequency) (14.24)

P = I

cos α(power) (14.25)

Quality factor (sharpness of resonance):

Q = ω

L/R (14.26)

Table 14.1 gives the analogues for electrical and mechanical quantities.

Table 14.1 Analogies between mechanical and electrical vibrations

Characteristic Mechanical Electrical

Inertia Mass (m) Inductance (L)

Stiffness Stiffness constant (k) Inverse capacitance (1/c)

Force Force (F)EMF

Resistance Frictional factor (r) Resistance (R)

Kinetic energy

Potential energy

Reactance mω −k/ω ωL − 1/ωC

Impedance





mω −







ωL −

ωC



Condition for oscillation r < 2

√

km R < 2

√

L/C

Resonance frequency

2π

√

k/m

2π

√

Quality factor ω

m/r ω

L/R

634 14 Electromagnetism II

Parallel Resonance Circuit

2π



−

(14.27)

When R = 0

2π

√

(14.28)

Maxwell’s Equations

Differential Form

∇. B = 0 (Gauss’ law of magnetostatics) (14.29)

∇. D = ρ (Gauss’ law of electrostatics) (14.30)

∇ × E =−

∂B

∂t

(Faraday’s law) (14.31)

∇ × H = J +

∂D

∂t

(Ampere–Maxwell law) (14.32)

Auxillary Equations

D = εE, D = ε

E + P (14.33)

B = μ

H, B = μ

(H + M) (14.34)

J = σ E, J = ρv (14.35)

where B = magnetic induction, ρ = charge density, E = electric ﬁeld, D = dis-

placement vector, H = magnetic intensity, J = current density, σ = conductivity,

M = magnetization, v = velocity, P = polarization.

The concept of displacement current can be explained by the working of a par-

allel plate capacitor placed in a vacuum and connected to a battery. As the capac-

itor gets charged current ﬂows through the wires but the usual current does not

pass between the plates of the capacitor plates. From considerations of continuity

Maxwell was led to postulate the existence of a displacement current equivalent to

the changing electric ﬁeld in the space between the plates.

On the theoretical side, Maxwell examined Ampere’s law, ∇×H = J, and

noticed that there is something strange about this equation. On taking the divergence

of this equation, the left-hand side will be zero, because the divergence of a curl is

always zero. This equation then requires that the divergence of J also be zero. But

if the divergence of J is zero, then the total ﬂux of current out of closed surface is

also zero.

14.1 Basic Concepts and Formulae 635

Now the ﬂux of current from a closed surface is the decrease of the charge inside

the surface. In general this cannot be zero because charges can be moved from one

place to another. This difﬁculty is avoided by adding the term ∂ D/∂t, where D =

E, on the right-hand side of (32).

Integral Form



B.ds = 0 (Gauss’ law for magnetostatics) (14.36)



E.ds =

enc

(Gauss’ law for electrostatics) (14.37)



E.ds =−

dϕ

(Faraday’s law of induction) (14.38)



B.ds = μ

dϕ

+ μ

enc

(Ampere-Maxwell law) (14.39)

Electromagnetic Waves

The E- and H-waves are transverse to the direction of propagation and are perpen-

dicular to each other.

∇×(∇×E) =−∇

E +∇(∇.E) (14.40)

Wave Equation

∇

E =−ω

E (14.41)

The Intrinsic Impedance

η =



(14.42a)



= 377  (free space) (14.42b)

The Poynting vector (S) represents the power in the electromagnetic wave and is

given by

S = E × H (14.43)

and points in the direction of propagation of the wave.

The skin thickness (δ) represents the depth to which an electromagnetic wave of

frequency f = ω/2π can penetrate a medium and is given by

636 14 Electromagnetism II

δ =



μσ ω

(14.44)

where μ is the permeability and σ is the conductivity.

Phase Velocity (v

) and Group Velocity (v

)

Phase velocity is just due to the nodes of the wave that are moving and not energy

or information. In fact the phase velocity can be greater than c, the velocity of light

= ω/k (14.45)

In order to know how fast signals will travel one must calculate the speed of

pulses or modulation caused by the interference of a wave of one frequency with

one or more waves of slightly different frequencies. The speed of the envelope of

such a group of waves is called the group velocity and is given by

= dω/dk (14.46)

Waveguides are hollow metallic structures in which the electromagnetic waves

are guided to travel from one place to another without much attenuation. Here, we

shall be concerned only with rectangular guide of cross-section of x = a and y = b ,

with the wave propagated in the z-direction.

−



mπ



−



nπ



(14.47)

where c is the velocity of light in free space, m and n are integers. Equation (14.47)

is valid both for TE

and TH

waves.

For the simplest case m = 1 and n = 0. For TE

wave.



1 −





(14.48)

where λ

is the free space wavelength.

= c



1 −





(14.49)

= c

(14.50)



1 −





(14.51)

14.2 Problems 637

14.2 Problems

14.2.1 The RLC Circuits

14.1 A2.5 μF capacitor is connected in series with a non-inductive resistor of

300  across a source of PD of rms value 50 V, alternating at 1000/2π Hz.

Calculate

(a) thermal values of the current in the circuit and the PD across the capacitor.

(b) the mean rate at which the energy is supplied by the source.

[Joint Matriculation Board of UK]

14.2 A3 resistor is joined in series with a 10 mH inductor of negligible resis-

tance, and a potential difference (rms) of 5.0 V alternating at 200/π Hz is

applied across the combination.

(a) Calculate the PD V

across the resistor and V

across the inductor.

(b) Determine the phase difference between the applied PD and the current.

[Joint Matriculation Board of UK]

14.3 An inductance stores 10 J of energy when the current is 5 A. Find its value.

14.4 A tuning circuit in a radio transmitter has a 4 × 10

−6

H inductance in series

with a 5 ×10

−11

F capacitance. Find

(a) the frequency of the waves transmitted.

(b) their wavelength.

14.5 A6 resistor, a 12  inductive r eactance and a 20  capacitive reactance

are connected in series to a 250 V rms AC generator. (a) Find the impedance.

(b) Estimate the power dissipated in the resistor.

14.6 At 600 Hz an inductor and a capacitor have equal reactances. Calculate the

ratio of the capacitive reactance to the inductive reactance at 60 Hz.

14.7 A capacitance has a reactance of 4  at 250 Hz. (a) Find the capacitance.

(b) Calculate the reactance at 100 Hz. (c) What is the rms current, if it is

connected to a 220 V 50 Hz line?

14.8 When an impedance, consisting of an inductance L and a resistance R in

series, is connected across a 12 V 50 Hz supply, a current of 0.05 A ﬂows

which differs in phase from that of the applied potential difference by 60

◦

Find the value of R and L. Find the capacitance of the capacitor which when

connected in series in the above circuit has the effect of bringing the current

into phase with the applied potential difference.

[University of London]

14.9 When a 0.6 H inductor is connected to a 220 V 50 Hz AC line, what is (a) the

rms current and (b) peak current?

638 14 Electromagnetism II

14.10 A simple alternator, when rotating at 50 revolutions/s, gives a 50 Hz alternat-

ing voltage of rms value 24 V. A 4.0  resistance R and a 0.01 H inductance

L are connected in series across its terminals. Assuming that the internal

impedance of the generator can be neglected, ﬁnd (a) the rms current ﬂowing;

(b) the power converted into heat; (c) the rms potential difference across each

component.

14.11 An AC circuit consists of only a resistor R = 100  and a source voltage

V = 0.5 V

at time t = 1/360 s. Assuming that at t = 0, V = 0, ﬁnd the

frequency.

14.12 Given that for a series LCR circuit the equation is

V = 0

If a similar equation is to be used for a parallel LCR circuit as in Fig. 14.1,

then show that

Fig. 14.1

14.13 Verify the equation c =

√

14.14 Show that in the usual notation the following combinations of physical quan-

tities have the units of time: (a) RC,(b)L/R,(c)

√

LC.

14.15 In an oscillating RLC circuit the amplitude of the charge oscillations drops

to one-half its initial value in 4 cycles. Show that the fractional decrement of

the resonance frequency is approximately given by

ω

= 0.00038.

14.16 Derive the equation for the current in a damped LC circuit for low damping.

14.2 Problems 639

14.17 Set up the equation for the RC circuit i n series and show that the input power

is the sum of the powers delivered to the inductor and capacitor.

14.18 Set up the equation for the RLC circuit in parallel and show that at any time

the Joule heat in the resistor comes from the energy stored in the inductor

and capacitor.

14.19 For the electrical circuit shown below when the switch K is closed the charge

and current are observed to oscillate (Fig. 14.2). Show that the differential

equation governing the charge of the system can be written as

+ 2γ

+ ω

Q = 0

where γ and ω

are to be determined in terms of the capacitance, C, induc-

tance, L, and resistance, R.

For a resistance R = 100 , capacitance C = 700 pF and inductance L =

80 mH calculate

(a) The natural frequency, f

, of the oscillation.

(b) The time constant, τ , for the decay.

Fig. 14.2

14.20 Solve the differential equation given in prob. (14.19) and obtain the time

period for damped harmonic motion.

14.21 A resistor, capacitor and inductor are connected in series across an AC volt-

age source shown as in Fig. 14.3.

(i) Find the magnitude of the inductive reactance X

of the inductor and

capacitive reactance X

of the capacitor.

(ii) Find the magnitude of the total impedance Z of the circuit and sketch

the impedance phasor diagram for this circuit.

(iii) Find the total current I

through the circuit.