• No results found

Examination paper for TFY4240 Electromagnetic theory

N/A
N/A
Protected

Academic year: 2022

Share "Examination paper for TFY4240 Electromagnetic theory"

Copied!
9
0
0

Laster.... (Se fulltekst nå)

Fulltekst

(1)

Department of Physics

Examination paper for

TFY4240 Electromagnetic theory

Academic contact during examination: Associate Professor John Ove Fjærestad Phone: 97 94 00 36

Examination date: 24 May 2018 Examination time (from-to): 9-13

Permitted examination support material: C Approved calculator

Rottmann: Matematisk Formelsamling (or an equivalent book of mathematical formulas)

Other information:

This exam consists of three problems, each containing several subproblems. In many cases it is possible to solve later subproblems even if earlier subproblems were not solved.

Under normal circumstances, each subproblem will be weighted approximately equally in the marking.

Some formulas can be found on the pages following the problems.

Language: English

Number of pages (including front page and attachments): 9

(2)

Problem 1

(a) Briefly describe the ”method of images” and the type of problems it can be used to solve.

· · ·

Let space be divided into two regions filled with simple dielectric media 1 and 2 with electric permittiv- ities 1 and 2 , respectively. The interface between the regions is an infinite plane. A point charge q sits in medium 2 a distance d from the interface.

We introduce a cartesian coordinate system with the z axis perpendicular to the interface and pass- ing through the point charge, with origin and direction chosen such that the interface is at z = 0, medium 1 is in the region z < 0, medium 2 is in the region z > 0, and the point charge has coordinates (x, y, z) = (0, 0, d) (see the figure).

𝑧

𝑞 𝑑

𝑧 = 0 medium 1

medium 2

(b) Use the method of images to find the potential V everywhere. [Hint: All image charges involved can be taken to be at distance d from the interface.]

(c) Find the force F on the point charge q. Comment on whether the result is reasonable for the special case 1 = 2 .

(d) Find the volume bound charge density.

(e) Find the surface bound charge density.

2

(3)

As illustrated in the figure below, an ohmic bar with resistance R, length L and mass m can slide without friction on two parallel, perfectly conducting rails (which extend infinitely to the right). A uniform magnetic field B points into the page. At time t = 0 a physicist sets the bar in motion by giving it an initial velocity v 0 towards the right. (For t > 0 the physicist does not act with any force on the bar.)

𝐿 𝑣 𝐵

(a) Argue that a current I will flow in the circuit consisting of the (moving) bar and the rails to the left of it. Find an expression for I which involves the bar’s velocity v towards the right. What is the direction of the current?

(b) Show that for t > 0 the time dependence of the current takes the form I(t) = I 0 exp(−αt), and give expressions for the constants α and I 0 . [Hint: Consider Newton’s 2nd law for the bar in the horizontal direction (i.e. in the bar’s direction of motion).]

(c) Calculate the total energy dissipated as Joule heat during the motion of the bar (i.e. from t = 0 to

t = ∞). Give a physical interpretation of the result.

(4)

Problem 3

By introducing the scalar potential V , it can be shown that the Maxwell equations for electrostatics are equivalent to a single differential equation for V , the Poisson equation ∇ 2 V = −ρ/ 0 , whose solution for a localized charge distribution is

V (r) = 1 4π 0

Z

d 3 r 0 ρ(r 0 )

|r − r 0 | . (1)

(a) By introducing the vector potential A, show that the Maxwell equations for magnetostatics are equivalent to a single differential equation for A. Derive this differential equation.

(b) (i) Show that B is invariant under the gauge transformation A → A 0 = A + ∇λ, where λ is an arbi- trary function of r. (ii) Argue that one can impose the so-called Coulomb gauge condition ∇ · A = 0 on A, and derive an expression for A(r) in the Coulomb gauge, for a (steady) localized current distribution.

(c) Show that for static problems, in the Coulomb gauge, the electromagnetic field momentum can be expressed as

P EM = Z

d 3 r ρ(r)A(r). (2)

[Hints: It may be useful to consider the cartesian components of P EM . You may assume that all ”surface terms” at infinity (obtained from integration by parts) vanish.]

(d) Consider a system consisting of three parts: A point charge q at (x, y, z) = (0, 0, d), a point charge

−q at (0, 0, −d), and a point magnetic dipole with moment m at the origin (0, 0, 0). Calculate P EM for this system.

(e) This question concerns the roles of the Poynting vector in general problems in classical electromag- netic theory. (i) Briefly state how the Poynting vector enters into calculations of electromagnetic field momentum. (ii) The Poynting vector also appears in energy considerations. Briefly describe the inter- pretation/role of the Poynting vector in this context.

4

(5)

Some formulas that you may or may not need (you should know the meaning of the symbols and possible limitations of validity):

2

∂V 2

∂n − 1

∂V 1

∂n = −σ f (1)

ρ b = −∇ · P , σ b = P · n ˆ (2)

ε = − dΦ

dt (3)

P = U I (4)

(a × b) i = ijk a j b k (5)

kij klm = δ il δ jm − δ im δ jl (6)

A dip (r) = µ 0

m × (r − r 0 )

|r − r 0 | 3 (7)

(6)

! "#$ $ "#% "&$ ' ( ) * ! + , - .

/ 0 1 2 3 4 5 6 , 7 89 :; < ( = * ! + , - >

? @A B C D E F G H ( * 9 "IJ K $ .

L M &N O P F - K , ! J ( = $ K 9 - $ ! ( Q .

R M S T U V V O W X Y J F , + + ( ) $Z 6 9 6 - $[ ( Q >

\ ] ^ _ ` a b c d c e f b g d a e f ` h b c d i j j k a e c l m n

o p q r s tu v w

x y z { + : Q | - ( + } ~ €  + ‚ Q | - ( + } sƒ „ … ( + | - ( + } †‡ ˆ G ‚Q } ‰

Š { + : Q | ‹ 8Q } Œ +  Q | + Ž Q }  „ - ( + | G ‚Q } †‘ „ - ( + } ‰

’ { - ( + | “” - ( + |  • + : Q | †–

— { y ˜ Š ™ „ ’ š sƒ › G œTQ | - ( + } ƒ ˜ + 8Q | G ‚Q }  „ - ( + | “ž

| $ , Q O F y ™ „ Š ’ . †Ÿ - ( + | - ( G } ž „ - ( + | + ‚ Q }    G ‚Q | “¡

} ¢£ $ , Q O F Š y . ¤ ¥ ¦ + ‚Q } § „ - ( + } 

¨ © ª «¬ ­ ® u v ¯

° y ± E - ( + } ° ƒ - ( + } ²³ • + "TQ } ¤

Š E + "&Q }  + ‚Q } ´µ „ - ( + } ¤

· ¸¹ º »¼ @ “ž

½

x E ¹ y „ Š ¾ ´¿ ¹ - ( + } ž ˜ + ‚Q } 

À } Á $ , Q  X F Š y . ¤ @Ã Ä + ÅTQ } Æ „ - ( + } 

 Ç

’ ’ “ È “

É Ê

Ë Ì

Í

Î Ï ÐÑ

Ò Ó

Ô Õ

(7)

×

Ø Ù Ú Û Ü Ý Þ ß Ù à Û á â Ú á ã Ý ä Ý Ü à å á æ ç è Ù é Û Ü Ú

ê ë ì í î ïï ðñ ò ó ô ë õ ö ÷ ø ñ

ù ú û ü ú ü ý þ ÿ ù ú þ ü ý

ò

ò

ò

! "

# $ % ò & $ % '

# ( )

* ô ì +ï +ë , - . öî ï / ñ

0 ü 1 ú 23245 ú 6 78 9 24ú ü þ ý ü : 2þ

; < = > > ? @ ò A B ò

C D E F G ê ê H I J C C K

L

# #

> ÷ M î ø M N ë Oñ ò P Q P R * S TU *

V ÷ , î ø M W X ÷ , Y î Z ë í . [ \ ò = ] ^ ‰ Š

‹

Œ



Ž 

(8)

’ “ ” • – — ˜ ’ ™ ” — š — ’ ›œ

 žŸ   ¡ ¢ £ ¤ ¥ ¦ § ¨ ©

ª « ¬ ­ ® ¯ ° ± ² ¬ ³´ µ ¶ · ² ± ­ ¬ ¸¹ ² º ª­ ± ° »

¯ ¼ ¬ ­ ± · ° ± ² ¬ ½¾ ° ¯ ­ º ² ¬ ¿ ² ¯ ­ À ° ¬

¢ £ ¤ Á ¦ § ¨ Â ¦   ¡ ©

· Ã Ä Å · Æ Ç Ä ÈÉ Æ · Å Ç ¬ Ê Ç ¯ Å Æ ¬

· Ë ¬ Å ¯ ­ À ° » ÌÉ ­ ± ¯ Å ± ° » Ê ° Í ¯ Å Í ­ ¬ Ê · ­ Î Ï » ° Ê ¯ ° Ð Å ¬ ­

Ñ

Ò Ó ¬ Å º ª Æ ­ ¬ Ô¾ Æ · Å Õ ­ ¬ Ö ­ Î · Å Æ ¬

· × Ø Å º Ù ­ ± ° Ä ÚÛ ° Ü · Å ± Ý ¬ Þ ­ Î · Å ± ° ¬

¯ ß ¬ Å ± ª Æ ­ ¬ ¸à Æ ¯ Å ± ­ Ä á ­ ± ¯ Ï Æ ¬

¯ â ¬ Å ± · ­ ± ° ¬ ³ã ¯ ° ¶ Ï ¬ ­ äå · ­ ® Å ¬ ° Ê ­ ¯ Å À ° Ä ä ° ª Å Î ­ ¬

æ ¡ § ¤ ç Á è ¡ £ é ê ë ¨ é ê ¡ ©

· ì Ø Å Õ · Å ± ­ Ä ¹í î

· ï î Ä Å ± ª Å Æ ¬ í³ î

· « « ¬ Å ± · Å ± ­ ¬ ¹Ô Å · Å Î ­ ¬ á ð ñ ­ ò ó ô ˜ õ ö ’ ™ š õ ÷ ” ø ’ • – ù ö ú

û

ü ý ë Á é¡ ç ¨ þ ÿ ¡ ¤ ý ¡ ºÀ · Å Æ ¬ Î   Ì Æ · Ä Þ Æ · ë ¬

è é ê ¡ £ ¡ § ¡ þ ÿ ¡ ¤ £ ¡ º® ª Å º Ý ¬ ¾ Ý ® ë

² ¦ ý   þ ÿ ¡ ¤ ý ¡ Õ® ¯ Å ± ­ ¬ ¶ ë ­ ¶

!

"

#

$

% &

'

(

(9)

* + , - . / 0 + 1 * 2 3 4 5 + 6

7 8 9 : ;< 9 = >? @ 9 A B @ 9 C DE FG 9 H ;I 9 = 9 A 9 C

J

K L M N OPQ R S TU V S WX YZ S [\ ] ^Z _ B ] `Z S Da

Z b Z c Z d

e fgh Q L i Q R j Q kT V l m n;

oZ

p q @ rZ p s @ tZ p u

Z b Z c Z v w

x y L z { V | } ~ Z p  Z p €  [‚ ƒ ~ Z p „ Z p   B ] ~ Z p € Z p q  Da

; … † ‡ ˆ † ‰ ˆ Š ‹

Œ

l Z c Z d Z d Z b Z b Z c

 Z Ž S Z  S Z  S

‘ ’ zM j fPM R ll “ S ;; ” ] ” ƒ … 

Z b Z c Z v

• – — ˜ ™ š › œ N : ; N L —\ ƒ ž N Ÿ  ¡ ] L ¢ £¤ Ÿ N ¥ ¦ l§ N ¨ © ž  ¢ £ ¤ Ÿ N L N Ÿ N ¥

K L M N flQ R S Pl V S ;; ªZ S «¬ ] ­® ¯Z S  ° ] ± ® ²Z S ¦

Z L L Z Ÿ L ¢ ³¤ Ÿ Z ¥

e fPh Q L i Q R j Q ´´ V T } µ; ¶ ® ` Z · L  p ¸ ¹ @ º » ^ Z ¼ ¢ ½¾¤ Ÿ p ¿ ¹ ] º ®

ÀZ

p Á

L Z L L  ½g¤ Ÿ Z Ÿ L ¢ £¤ Ÿ Z ¥

à y L Ä U{ Å | } ; º Æ Ç È Z · ¢ £¤ Ÿ p Á ¹ †

‡Z

É ¿ Ê «‚

L ¢ ½P¤ Ÿ Z Ÿ Z ¥

] ËÌ Í Î Ì

ÏZ

p ¸ Š Ð Z · L p Á ¹ Ñ  Ò @ Ó® Í ` Z · L p ¿ Ô Õ …Z

p ¸ Ö ¦

L ¢ ½P¤ Ÿ Z ¥ Z L L Z L Z Ÿ

× z l Ø  Ù Ú ~  Z S  Û Z ~ l Ÿ Z _ Ü ® Z Ž S

’ M j fM R lP S Ý Þ ß L ß ] à  l Ÿ á ¢ ½¤ á ] ⠍ P  Ÿ ãZ ¥ 

L L L L ¢ ½¤ L ¢ ä ¤

¡

7 æ ç è « é™ š ê ë N ê ;; N ì í\ ] î N ] N d Dð ¾ñ N ¨ XX î N î N ò d

Referanser

RELATERTE DOKUMENTER

The cost of using force to secure national interests in the near abroad may increase significantly if economic growth is hampered and/or Russia’s role in international

The system can be implemented as follows: A web-service client runs on the user device, collecting sensor data from the device and input data from the user. The client compiles

In April 2016, Ukraine’s President Petro Poroshenko, summing up the war experience thus far, said that the volunteer battalions had taken part in approximately 600 military

This report documents the experiences and lessons from the deployment of operational analysts to Afghanistan with the Norwegian Armed Forces, with regard to the concept, the main

The AUTODYN-2D simulations have been found to give results that are in good agreement with the experiment, whereas the cavity expansion theory shows poor agreement with the

However, a shift in research and policy focus on the European Arctic from state security to human and regional security, as well as an increased attention towards non-military

Scholars from different social sciences devoted to educational research (educationalists, economists, political scien- tists, sociologist and so) shared their insight and

Although, particularly early in the 1920s, the cleanliness of the Cana- dian milk supply was uneven, public health professionals, the dairy indus- try, and the Federal Department