Problem 1-3 Wangsness Electromagnetic FieldsFind the relative vector ~R of the point P (2;�2; 3) with respectto P 0(�3; 1; 4).Solution:The vector ~R is given by:~R = ~P � ~P 0 = (2 � [�3]; [�2]� 1; 3� 4) = (5; � 3; � 1)= 5 x� 3 y � z

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Problem 1-7 Wangsness Electromagnetic FieldsShow that ~A � ( ~B � ~C) equals the volume of a parallelepipedwhose sides at a common corner are parallel to the vectors ~A, ~Band ~C.Solution:The vector product ~B� ~C is numerically equal to the product of the lengthsB and C and the sine of the angle between them. This is equal to the baseB times height A sin �AB of the wall which they bound. This is just the areaof this wall which we shall refer as the \base" of the parallelepiped.Dotting this with ~A gives this area times the length of A times the cosineof the angle between ~A and the normal to this base. So the overall productis the base � height of the parallelepiped which in turn is just the volume.

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Problem 1-8 Wangsness Electromagnetic FieldsA family of hyperbolae in the xy plane is given by u = xy.a) Find ~ru.b) Find the component of the vector ~A = 3 x + 2 y + 4 z in thedirection of ~ru at the point on the curve for which u = 3 andx = 2.Solution:a) ~ru = @u@x x+ @u@y y + @u@z zwhere u = xy. Therefore ~ru = y x+ x yb) The component of the vector ~A = 3 x+ 2 y + 4 z in the direction of~ru is ~A � ~ru= ���~ru��� = 3y + 2xpx2 + y2When u = 3 and x = 2 then xy = 2y = 3 so that x = 2 and y = 3=2.Therefore ~A � ~ru = 17=5:

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Problem 1-9 Wangsness Electromagnetic FieldsThe equation giving a family of ellipsoids is:u = x2a2 + y2b2 + z2c2Find the unit vector normal to each point of the surface of theseellipsoids.Solution:The unit normal to a surface u(~r) =constant isn = ~ru���~ru���In the present case: ~ru = 2� xa2 x + yb2 y + zc2 z�Therefore the normal at point (x; y; z) isn = � xa2 x + yb2 y + zc2 z� x2a4 + y2b4 + z2c4!� 12

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Problem 1-13 Wangsness Electromagnetic FieldsA vector �eld is given by~A(~r) = xy x+ yz y + zx za) Evaluate the ux through the rectangular parallelepiped sidesa; b; c respectively in the x; y; z dimensions and with corner atthe origin and in the �rst octant.b) Evaluate R ~r� ~Ad� over the volume of this same parallelepipedand compare your result with that from part a).Solution:a) The ux through the rectangular parallelepiped isZS ~A � d~awhere S is the whole surface. For the side I at x = 0 the vectord~a = �dydzx, for the side at y = 0 then d~a = �dxdzy and for theside at z = 0 then d~a = �dxdyz. The sides opposite these have thesevectors reversed in sign.Therefore the surface integral isZS ~A � d~a = Z b� dy Z c� dz xy�����x=a � Z b� dy Z c� dz xy�����x=0 +Z c� dz Z a� dx yz����y=b � Z c� dz Z a� dx yz����y=0 +Z a� dx Z b� dy zx�����z=c � Z a� dx Z b� dz zx�����z=0= 12 �ab2c+ abc2 + a2bc�5

b) The divergence of ~A is y+z+x. Therefore the integral over the volumeof the parallelepiped V isZV ~r � ~Ad� = Z a� dx Z b� dy Z c� dz(y + z + x)= 12 �a2bc+ ab2c+ abc2�This is the same result as in a) as we would expect from the divergencetheorem.

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Problem 1-15 Wangsness Electromagnetic FieldsA vector �eld is given by~A(~r) = x2y x+ xy2 y + a3e��y cos�x zwhere a; � and � are constants (independent of x; y; z).a) Evaluate the line integral over the closed loop C in the planez = 0 along the curve y2 = kx from x = 2 to x = 0, then alongthe y axis back to the y at which x = 2 and then at constanty back to the starting point.b) Evaluate RS(~r � ~A) � d~a over the at surface enclosed by theloop C and compare your result with that from part a).Solution:a) The line integral along the curved part (y2 = kx) is �rst. We havey2 = kx ; 2ydy = kdx ; dy = 12dxskxTherefore this integral isZ x=�x=2 (Ax dx + Ay dy) = Z x=�x=2 �x2y dx + xy2 dy�= pk Z �2 dx �x5=2 + 12kx3=2�= k 12 �27x 72 + 15kx 52 ��2 = �p2k�167 + 45k�Along the y axis the integral is R Aydy at x = 0 which is zero.The part of the integral parallel to the x axis is at y = p2k and isZ 2� Axdx = Z 2� x2ydx = p2k83The path integral around C is thereforep2k� 821 � 45k�7

b) The curl of ~A is~r� ~A = ��a3e��y cos�x x + �a3e��y sin�x y + (y2 � x2) zThe vector d~a is�dxdyz so that the integral of the curl over the surfaceisZS(~r� ~A) � d~a = � Z 2� dx Z p2kpkx dy (y2 � x2)= � Z 2� dx �p2k �pkx� �k3 (2� x)� x2�= p2k Z 2� dx �1�rx2��x2 + k3x� 2k3 �= p2k �13x3 + k6x2 � 2k3 x�rx2 �27x3 + 2k15x2 � 4k9 x��2�= p2k �83 + 2k3 � 4k3 � 167 � 8k15 + 8k9 �= p2k � 821 � 4k5 �which agrees with part a) as expected.

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Problem 1-19 Wangsness Electromagnetic FieldsA vector �eld is given by~A(~r) = a �+ b �+ c zwhere a; b and c are constants (independent of �; � and z).a) Is ~A a constant vector?b) Evaluate ~r � ~A.c) Evaluate ~r� ~A.d) Find ~A in Cartesian coordinates with components in termsof x; y and z.e) Find ~A in spherical polar coordinates with components interms of r; � and �.Solution:a) The vector changes direction as the position is changed because � and� change their direction. The magnitude of ~A however does remainconstnt.b) The divergence of ~A is~r � ~A = 1� @@� (�A�) + 1� @A�@� + @Az@z = a�NOTE - this is not zero as it would have been were ~A a constantvector.c) The curl of ~A is~r� ~A = ��1� @Az@� � @A�@z �+ ��@A�@z � @Az@� �+ z �1� @@� (�A�)� 1� @A�@� �= z b�NOTE - this also is not zero as it would have been were ~A a constantvector. 9

d) The unit vectors are related by:� = cos� x+ sin� y ; � = � sin� x+ cos� y ; z = zTherefore our vector can be written as~A = (a cos�� b sin�) x+ (a sin�+ b cos�) y + c zEliminate � usingcos� = xpx2 + y2 ; sin� = ypx2 + y2to obtain: ~A = (ax� by) x+ (ay + bx) ypx2 + y2 + c ze) The unit vectors in spherical coordinates are related to those for cylin-drical coordinates by: � = sin � r + cos � �� = �z = cos � r � sin � �Therefore our vector can be written as~A = (a sin � + c cos �) r + (a cos � � c sin �) � + b �

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Problem 1-21 Wangsness Electromagnetic FieldsFind the divergence ~r � r for the position vector ~r in rectangular,cylindrical and spherical polar coordinates and show that it is thesame in all these.Solution:In rectangular coordinates~r = x x+ y y + z z~r � ~r = @rx@x + @ry@y + @rz@z = 3In cylindrical coordinates~r = � �+ z z~r � ~r = 1� @@� (�r�) + 1� @r�@� + @rz@z = 2 + 1 = 3In spherical polar coordinates~r = r r~r � ~r = 1r2 @@r �r2rr�+ 1r sin � @@� (sin �r�) + 1r sin � @r�@� = 3Clearly all three coordinates give the same result.

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Problem 1-23 Wangsness Electromagnetic Fieldsa) For a vector �eld ~A = 4 r+3 ��2 � �nd the line integral counterclockwise around a plane loop in the xy plane consisting ofthe �rst quadrant of a circle radius r� centred at the originand the portions of the x and y axes inside this quarter circle.b) Find the surface integral of ~r� ~A over the xy plane enclosedby the loop.Solution:a) The path around the circular arc has path d~r = rd� �. So the lineintegral for this arc isZ ~A � d~r = Z �=2� (�2)r�d� = � �r�For the path along the y axis heading towards the origin the path isd~r = dr r So the line integral for this arc isZ ~A � d~r = Z �r� 4dr = � 4r�For the path along the x axis heading away from the origin the pathis d~r = dr r So the line integral for this arc isZ ~A � d~r = Z r�� 4dr = 4r�Therefore the path integral around the whole loop isZC ~A � d~r = ��r�b) The curl of ~A is~r � ~A= rr sin � � @@� (sin �A�)� @A�@� �+ �r � 1sin � @Ar@� � @@r (rA�)�+ �r � @@r (rA�)� @Ar@� �= 2r �� cot � r + � + 32 ��12

The surface di�erential isd~a = �r2 sin �d�d� r � r sin �drd� � � rdrd� �Then the product~r� ~A � d~a = 2��r cos �d�d�+ sin �drd�+ 32drd��Since we work in the xy plane, � = �=2 and only the � componentof the surface di�erential is non zero. Given the counter clockwisecirculation around the loop, this is the negative value �rdrd�. SoZS(~r� ~A) � d~a = �2 Z r�� dr Z �=2� d� = � �r�which is the same as the line integral as expected from Stokes' theorem.

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Problem 1-25 Wangsness Electromagnetic Fieldsa) Apply the divergence theorem to a constant, though arbi-trary, vector �eld ~A(~r) to show that the total vector area R d~aaround a closed surface is zero.b) Using such a �eld show also that for any closed path thevector sum R d~r = 0.Solution:a) Let ~A(~r) = a x + b y + c z (where a; b and c are independent of x; yand z) be the constant but arbitrary �eld.Since all derivatives of a; b and c with respect to x; y and z are zero,the divergence of ~A is also zero everywhere.Apply the divergence theorem to a volume V enclosed by a closedsurface S to obtain ZS ~A � d~a = ZV (~r � ~A)d� = 0 soa ZS dax + b ZS day + c ZS daz = 0for all a; b and c. This can only be so if each of the integrals is inde-pendently zero. ThereforeZS d~a = ZS dax x+ ZS day y + ZS daz z = 0 Q.E.D.b) Since all derivatives of a; b and c with respect to x; y and z are zero,the curl of ~A is also zero everywhere.Applying Stokes' theorem to a closed loop C anywhere and an arbi-trary surface S bounded by this loop, we see that the path integralZC ~A � d~r = ZS(~r� ~A) � d~a = 0 soa ZC drx + b ZC dry + a ZC drz = 0for arbitrary a; b or c. Therefore the integrals are again separately zeroand thereforeZC d~r = ZC drx x+ ZC dry y + ZC drz z = 0 Q.E.D.14