Solution Electromagnetic Field Wangsness Chap 19 Selected number

Electricity and Magnetisme Homework V Solution Roald K. Wangsness, Electromagnetic Fields 2nd Edition, Chapter 19 by  Chalis Setyadi  1.  Exercise 19-15  : A point dipole  m  is located at the origin, but it has no special orientatio orientation n with respect to the coordinate coordinate axes. (For (For example,m is not parallel to any of the axes.) Express its potential  A  at a point  r  in rectangular coordinates, and find the rectangular components of  B . Show that B  can be written in the form B(r) =

µ0  [3 (m · ˆr) ˆr − m] 4πr 3

(19 − 55)

and compare with (8-84). Solution  : In rectangular coordinate, we can write :

ˆ + my y ˆ  +  mz zˆ = mx x ˆ + y y ˆ  + z z ˆ r = xx

m

r

= |r| = x2 + y 2 + z 2





1/2

(1)

Then inserting equation equation (1) to equation (19-55) we get B(r)

= =

B(x , y , z )

=

µ0  [3 (m · ˆr) ˆr − m] 4πr 3 m µ0 3 (m · r) r − 3 5 4π r r µ0 4π

 



ˆ + y y ˆ  +  z zˆ) 3 (mx x + my y +  mz z ) (xx 5/2

(x2 + y 2 + z 2 )

−

ˆ + my y ˆ  + mz zˆ)  ( mx x 3/2

(x2 + y 2 + z 2 )



(2) And we can write Bx

=

µ0 4π

By

=

µ0 4π

Bz

=

µ0 4π

  

3 (mx x + my y +  mz z ) x 5/2

(x2 + y 2 + z 2 )

3 (mx x + my y +  mz z ) y (x2 + y 2 + z 2 )

5/2

3 (mx x + my y +  mz z ) z 2 5/2

(x2 + y 2 + z )

− − −

mx

(x2 + y 2 + z 2 )

3/2

my

(x2 + y 2 + z 2 )

3/2

mz

(x2 + y 2 + z 2 )3/2

  

(3)

Then we will show that we can get expressions in equation ( 3) from B  = ∇ × A  in rectangular coordinat. The first problem problem in this exercise is expressing expressing potential potential vector vector  A  in rect-

1

angular coordinat. We can can do it using the terms in equation ( 1) µ0 m × r 4π r 3 ˆ + my y ˆ  +  mz ˆz) × (xx ˆ + y y ˆ  +  z ˆz) µ0 (mx x 3 2 / 4π (x2 + y 2 + z 2 )

=

A

=

(4)

µ0 (my z − mz y) ˆ x + (mz x − mx z ) ˆ y + ( mx y − my x) ˆ z 3 / 2 4π (x2 + y 2 + z 2 )

=

(5)

Using expression A  in equation (4) and vector triple product in equation (1-121) in the book, we can write B =

B

=





r µ0 m × r µ0 ∇ × A = ∇ ×  = ∇× m× 3 3 4π r 4π r µ0 r r r r ∇ · 3 m − (∇ · m) 3 + 3 · ∇ m − (m · ∇) 3 r r r r 4π

 





 







 (6)

  

We know that the second term (∇ · m) r  and the third term r · ∇ m of equation (6) are zero (0) because m  is constant (not a function of any variable) as it can be seen from equation (1). The fisrt term is also zero because ∇ · r  = 0 as we can check it in rectangular coordinat: r

3

r

3

r

3

∇·

r r3

=



=

∂  ∂x

+

∂  ∂z

=



       

∂  ∂  ∂  ˆ +  ˆ ˆz x y + ∂x ∂y ∂z

 

x

·

2 3/2

(x2 + y 2 + z ) z

ˆ + y y ˆ  +  z ˆz xx

(x2 + y 2 + z 2 )3/2

 +

∂  ∂y

y

(x2 + y 2 + z 2 )3/2

3/2

(x2 + y 2 + z 2 ) 3

(x2 + y 2 + z 2 )3/2

−

3 x2 + y 2 + z 2

(x2 + y 2 + z 2 )5/2

=0

(7)

Then the remaining term of equation (6) is the last term B

=

r µ0 − (m · ∇) 3 4π r

=

µ0 − 4π

 



∂  ∂  ∂  + mz mx  +  my ∂x ∂y ∂z

We can solve equation (8) using two methods.

2



ˆ + y y ˆ  +  z zˆ xx (x2 + y 2 + z 2 )

3/2



(8)

(a)   Direct calculation in rectangular coordinate  For the x  component of  B = −

Bx

µ0 4π

µ0 4π

=





mx

∂  ∂  ∂   +  my + mz ∂x ∂y ∂z

3x (mx x + my y +  mz z )

−

(x2 + y 2 + z 2 )5/2





x

(x2 + y 2 + z 2 ) mx

(x2 + y 2 + z 2 )3/2

3/2



(9)

The y  and  z  component of  B can be calculated using the same method By

=

µ0 4π

Bz

=

µ0 4π

 

3y (mx x + my y +  mz z ) (x2 + y 2 + z 2 )5/2 3z (mx x + my y +  mz z ) 2 5/2

(x2 + y 2 + z )

− −

my

(x2 + y 2 + z 2 )3/2 mz

(x2 + y 2 + z 2 )

3/2

 

 (10)

Comparing equation (9) and (10) with equation (3), we can see that thats all are the same expression. QED. (b)   Using tensor notation  We can write equation (8) in tensor notation as B

= −

r µ0 (m · ∇) 3 r 4π

= −

µ0 4π

= −

µ0 mi 4π

                       mi

=

µ0 mi − 4π

=

µ0 4π

=

µ0 xj eˆj 4π

= =

xj eˆj (Σx2j )3/2

∂  ∂x i

∂x j eˆj ∂x i (Σx2j )3/2

 + xj eˆj

 + xj eˆj

eˆj (Σx2j )3/2

 + xj eˆj

3mj xj (Σx2j )5/2

1 (Σx2j )3/2

∂x

eˆj δ ij (Σx2j )3/2

−mi δ ij

∂  ∂x i

−

−3xj ∂xji

(Σx2j )5/2

3mi xj δ ij (Σx2j )5/2

mj eˆj (Σx2j )3/2

µ0 (3m · r) r m − 3 5 r r 4π µ0  [(3 m · ˆr) ˆr − m] QED 4πr 3

(11)

2.   Exercise 19-10  : A point dipole m1  is located at r1  and another point dipole m2  is at r2 . Either by direct calculation or by transcription of the corresponding results for the electric case, show that the potential energy 3

of  m2   in the induction of  m1   is given by   the dipole-dipole interaction  energy  µ0 ˆ U DD  = (m1 · m2 ) − 3 m1 · R 4πR 3





   ˆ m2 · R

(19 − 56)

where R  =  r 2 − r1 . Similarly, find the force F2 on m2 .

Solution  : Magnetic field produced by m1  in position m2  can be calcu-

lated from equation (19-55) B1  =

µ0 ˆ  ˆ 3 m1 · R R − m1 4πR 3

 

with



 

(12)

= r2 − r1

R

|R| = =

 

(x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 x2 + y 2 + z 2



1/2

1/2



(13)

Then the potentioal energy of  m2 is 

= −m2 · B1

U DD

= =

µ0 ˆ  ˆ m2 · m1 − 3 m1 · R R 4πR3 µ0 ˆ ˆ (m1 · m2 ) − 3 m1 · R m2 · R 4πR3

and the force on m2 is F2





    

 

(14)



= −∇U DD  = ∇ (m2 · B1 ) = B1 × (∇ × m2 ) + m2 × (∇ × B1 ) + ( B1 · ∇) m2  + ( m2 · ∇) B1 (15)

The first and third term of equation ( 15) are zero because  m 2  is contant. The second term is also zero because the source of radiation B1  is in m1 but our integral is over  m 2 , so there are no source of magnetic field ( J  = 0) enclosed by the integral

 

(∇ × B1 ) · da2  =

 

B1 · ds2  =  µ 0 J = 0

(16)

Then the remaining term for F2 is F2

= ∇ (m2 · B1 ) = (m2 · ∇) B1 = =



∂  ∂  ∂  + m2y + m2z m2x ∂x 2 ∂y 2 ∂z 2

µ0 4π



m2x

        µ0 ˆ  ˆ 3 m1 · R R − m1 4πR 3

∂  ∂  ∂  + m2y + m2z ∂x 2 ∂y 2 ∂z 2

There are two methods to solve equation (17). 4

3 (m1 · R) R 5

R

−

m1 R3

(17)

(a)  Direct calculation in Cartesian coordinat  The  x  component of  F2 are =

F 2x

= + + = + −







3x (m1x x + m1y y +  m1z z ) m1x − 2 2 2 2 5/2 (x + y + z ) (x + y 2 + z 2 )3/2 µ0 3 (2xm1x  +  ym 1y  +  zm 1z )  15 x2 (xm1x  +  ym 1y  +  zm 1z ) 3xm1x m2x − + 2 2 2 2 5/2 2 2 2 7/2 4π (x + y + z ) (x + y + z ) (x + y 2 + z 2 )5/2 µ0 3xm1y  15 xy (xm1x  +  ym 1y  +  zm 1z ) 3ym 1x + 2 m2y − 2 2 2 5/2 2 2 2 7/2 4π (x + y + z ) (x + y + z ) (x + y 2 + z 2 )5/2 µ0 3xm1z  15 xz (xm1x  +  ym 1y  +  zm 1z ) 3zm 1x − + 2 m2z 2 2 2 5/2 2 2 2 7/2 4π (x + y + z ) (x + y + z ) (x + y 2 + z 2 )5/2 µ0 3m1x (xm2x  +  ym 2y  +  zm 2z )  3 m2x (xm1x  +  ym 1y  +  zm 1z ) + 4π (x2 + y 2 + z 2 )5/2 (x2 + y 2 + z 2 )5/2 µ0 3x(m1x m2x  +  m1y m2y  +  m1z m2z ) 4π (x2 + y 2 + z 2 )5/2 µ0 15x(xm2x  +  ym 2y  +  zm 2z )(xm1x  +  ym 1y  +  zm 1z )   (18) 4π (x2 + y 2 + z 2 )7/2 µ0 4π

∂  ∂  ∂  m2x  +  m2y + m2z ∂x ∂y ∂z

  

  



 





Using the same method,  y and z  components of  F2 are F 2y

F 2z

=

µ0 4π

+

µ0 4π

−

µ0 4π

=

µ0 4π

+

µ0 4π

−

µ0 4π

     



3m1y (xm2x  +  ym 2y  +  zm 2z )  3 m2y (xm1x  +  ym 1y  +  zm 1z ) + (x2 + y 2 + z 2 )5/2 (x2 + y 2 + z 2 )5/2 3y(m1x m2x  +  m1y m2y  +  m1z m2z ) (x2 + y 2 + z 2 )5/2 15y (xm2x  +  ym 2y  +  zm 2z )(xm1x  +  ym 1y  +  zm 1z ) (x2 + y 2 + z 2 )7/2 3m1z (xm2x  +  ym 2y  +  zm 2z )  3 m2z (xm1x  +  ym 1y  +  zm 1z ) + (x2 + y 2 + z 2 )5/2 (x2 + y 2 + z 2 )5/2 3z (m1x m2x  +  m1y m2y  +  m1z m2z ) (x2 + y 2 + z 2 )5/2 15z (xm2x  +  ym 2y  +  zm 2z )(xm1x  +  ym 1y  +  zm 1z )   (19) (x2 + y 2 + z 2 )7/2











Then F2

ˆ + F 2y y ˆ  + F 2z ˆz = F 2x x 3µ0 ˆ + (m1 · R ˆ )m2  + ( m2 · R ˆ )m1 − 5(m1 · R ˆ )(m2 · R ˆ )R ˆ = (m1 · m2 )R 4πR 6 (20)





(b)   Using tensor notatation 

5



We can write equation (17) as F2



m1k eˆk (3m1j xj )xk eˆk − 2 5/2 (Σxk ) (Σx2k )3/2

=

µ0 ∂  m2i ∂x i 4π

=

µ0 ∂  3m2i m1j ∂x i 4π

=

µ0 3m2i m1j 4π

−

µ0 ∂  m2i m1k ∂x i 4π

=

µ0 3m2i m1j 4π

=

µ0 4π

= =



 

 



xj xk (Σx2k )5/2





∂  eˆk − m2i m1k ∂x i



1 (Σx2k )3/2

∂x j xk xj ∂x k ∂  + + xj xk 2 5/2 2 5/2 ∂x i (Σxk ) ∂x i ∂x i (Σxk )



1 (Σx2k )3/2





 eˆk

1

(Σx2k )5/2



eˆk

eˆk

xk xj  x j xk (−5xk δ ki ) + δ ji δ   + 2 5/2 2 5/2 ki (Σxk ) (Σxk ) (Σx2k )7/2





eˆk −

−3xk δ ki µ0  ˆ m2i m1k ek 4π (Σx2k )5/2

3m2i m1i xk  3 m2k m1j xj  15 m2i m1j xj xi xk µ0 3m2i m1k xi eˆk  + eˆk + − 2 5/2 2 5/2 2 7/2 4π (Σx2k )5/2 (Σxk ) (Σxk ) (Σxk ) 3(m1 · m2 )R  3( m1 · R)m2  15( m1 · R)(m2 · R)R  3( m2 · R)m1 + − + 5 5 7 5

µ0 R R R R 4π 3µ0 ˆ + (m1 · R ˆ )m2  + ( m2 · R ˆ )m1 − 5(m1 · R ˆ )(m2 · R ˆ )R ˆ (m1 · m2 )R 4πR 6



3.   Exercise 19-12   : A circular ring of radius a  lies in the xy   plane with the origin at its center. It carries a current I   circulating clockwise when viewed from a point on the positive z  axis. A point dipole m  =  m zˆ  is on the  z  axis. Find the  z  component of the force on  m . Solution  : Magnetic field produced by  I  in the dipole m  position is µ0 Ids × R   (22) B  = 4π C  R3

 

with R

ˆ + y y ˆ ) =  z ˆz − (a cos φx ˆ + a sin φy ˆ) = z ˆz − (xx 1/2

= a2 + z 2 ˆ + dy y ˆ  = −a sin φdφx ˆ + a cos φdφy ˆ ds = dxx ˆ + z sin φy ˆ  + azˆ) dφ   ds × R = a (z cos φx



R



(23)

and because the current  I  circulates clockwise, it has negative sign, so B

=

µ0 Ia − 4π

= −

2π

  0

ˆ + z sin φy ˆ  +  aˆz z cos φx dφ (a2 + z 2 )3/2

µ0 Ia2

2 (a2 + z 2 )

3/2

ˆz

 

(24)

The potential energi of dipole m  is U  =

−m · B

= mˆz ·

µ0 Ia2

ˆ  = z 3/2

2 (a2 + z 2 ) 6

µ0 mIa2

2 (a2 + z 2 )3/2

 

(25)



  

(21)

Then the force on m  is F

= −∇U  = −∇ =



3µ0 mIa 2 z 5/2

2 (a2 + z 2 )

7

µ0 mIa 2

2 (a2 + z 2 )3/2

ˆ z

  

(26)