# Hayt8e_SM_Ch5.pdf

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CHAPTER 5 5.1. Given the current density J = −104 [sin(2x)e−2y ax + cos(2x)e−2y ay ] kA/m2 : a) Find the total current crossing the plane y = 1 in the ay direction in the region 0 x 1, 0 z 2: This is found through Z Z Z 2Z 1 Z 2Z 1 Ø Ø Ø Ø I= J · nØ da = J · ay Ø dx dz = −104 cos(2x)e−2 dx dz S y=1 S 0 0 0 0 Ø1 1 Ø = −104 (2) sin(2x)Ø e−2 = −1.23 MA 2 0 b) Find the total current leaving the region 0 x, x 1, 2 z 3 by integrating J·dS over the surface of the cube: Note first that
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CHAPTER 55.1.  Given the current density  J  = − 10 4 [sin(2 x ) e − 2 y a x  + cos(2 x ) e − 2 y a y ]kA / m 2 :a) Find the total current crossing the plane  y  = 1 in the  a y  direction in the region 0  < x <  1,0  < z <  2: This is found through I   =    S  J · n  S  da  =    20    10 J · a y  y =1 dxdz  =    20    10 − 10 4 cos(2 x ) e − 2 dxdz = − 10 4 (2)12 sin(2 x )  10 e − 2 = − 1 . 23MAb) Find the total current leaving the region 0  < x,x <  1, 2  < z <  3 by integrating  J · dS  overthe surface of the cube: Note ﬁrst that current through the top and bottom surfaces willnot exist, since  J  has no  z  component. Also note that there will be no current through the x  = 0 plane, since  J  x  = 0 there. Current will pass through the three remaining surfaces,and will be found through I   =    32    10 J ·  ( − a y )  y =0 dxdz  +    32    10 J ·  ( a y )  y =1 dxdz  +    32    10 J ·  ( a x )  x =1 dydz = 10 4    32    10  cos(2 x ) e − 0 − cos(2 x ) e − 2   dxdz − 10 4    32    10 sin(2) e − 2 y dydz = 10 4  12  sin(2 x )  10 (3 − 2)  1 − e − 2  + 10 4  12  sin(2) e − 2 y  10 (3 − 2) = 0c) Repeat part  b , but use the divergence theorem: We ﬁnd the net outward current throughthe surface of the cube by integrating the divergence of   J  over the cube volume. We have ∇ · J  =  ∂  J  x ∂  x  +  ∂  J  y ∂  y  = − 10 − 4  2cos(2 x ) e − 2 y − 2cos(2 x ) e − 2 y   = 0 as expected 5.2.  Given  J  =  − 10 − 4 ( y a x  +  x a y )A / m 2 , ﬁnd the current crossing the  y  = 0 plane in the  − a y direction between  z  = 0 and 1, and  x  = 0 and 2.At  y  = 0,  J ( x, 0) = − 10 4 x a y , so that the current through the plane becomes I   =    J ·  d S  =    10    20 − 10 4 x a y  ·  ( − a y ) dxdz  = 2 × 10 − 4 A58  5.3.  Let J  = 400sin θ r 2 + 4  a r  A / m 2 a) Find the total current ﬂowing through that portion of the spherical surface  r  = 0 . 8,bounded by 0 . 1 π  <  θ  <  0 . 3 π , 0  <  φ  <  2 π : This will be I   =     J · n  S  da  =    2 π 0    . 3 π . 1 π 400sin θ ( . 8) 2 + 4( . 8) 2 sin θ d θ d φ  = 400( . 8) 2 2 π 4 . 64    . 3 π . 1 π sin 2 d θ = 346 . 5    . 3 π . 1 π 12[1 − cos(2 θ )] d θ  = 77 . 4Ab) Find the average value of   J  over the deﬁned area. The area isArea =    2 π 0    . 3 π . 1 π ( . 8) 2 sin θ d θ d φ  = 1 . 46m 2 The average current density is thus  J avg  = (77 . 4 / 1 . 46) a r  = 53 . 0 a r  A / m 2 . 5.4.  If volume charge density is given as  ρ v  = (cos ω t ) /r 2 C / m 3 in spherical coordinates, ﬁnd  J . Itis reasonable to assume that  J  is not a function of   θ  or  φ .We use the continuity equation (5), along with the assumption of no angular variation towrite ∇ · J  = 1 r 2 ∂ ∂  r  r 2 J  r   = − ∂ρ v ∂  t  = − ∂ ∂  t  cos ω tr 2   =  ω sin ω tr 2 So we may now solve ∂ ∂  r  r 2 J  r   =  ω sin ω t by direct integration to obtain: J  =  J  r a r  =  ω sin ω tr  a r  A / m 2 where the integration constant is set to zero because a steady current will not be createdby a time-varying charge density.59  5.5.  Let J  = 25 ρ  a ρ −  20 ρ 2 + 0 . 01  a z  A / m 2 a) Find the total current crossing the plane  z  = 0 . 2 in the  a z  direction for  ρ  <  0 . 4: Use I   =    S  J · n  z = . 2 da  =    2 π 0    . 40 − 20 ρ 2 +  . 01 ρ d ρ d φ = −  12  20ln( . 01 +  ρ 2 )  . 40 (2 π ) = − 20 π ln(17) = − 178 . 0Ab) Calculate  ∂ρ v / ∂  t : This is found using the equation of continuity: ∂ρ v ∂  t  = −∇ · J  = 1 ρ∂ ∂ρ ( ρ J  ρ ) +  ∂  J  z ∂  z  = 1 ρ∂ ∂ρ (25) +  ∂ ∂  z   − 20 ρ 2 +  . 01   = 0c) Find the outward current crossing the closed surface deﬁned by  ρ  = 0 . 01,  ρ  = 0 . 4,  z  = 0,and  z  = 0 . 2: This will be I   =    . 20    2 π 0 25 . 01 a ρ  ·  ( − a ρ )( . 01) d φ dz  +    . 20    2 π 0 25 . 4 a ρ  ·  ( a ρ )( . 4) d φ dz +    2 π 0    . 40 − 20 ρ 2 +  . 01 a z  ·  ( − a z ) ρ d ρ d φ  +    2 π 0    . 40 − 20 ρ 2 +  . 01 a z  ·  ( a z ) ρ d ρ d φ  = 0since the integrals will cancel each other.d) Show that the divergence theorem is satisﬁed for  J  and the surface speciﬁed in part  b .In part  c , the net outward ﬂux was found to be zero, and in part  b , the divergence of   J was found to be zero (as will be its volume integral). Therefore, the divergence theoremis satisﬁed. 5.6.  In spherical coordinates, a current density  J  =  − k/ ( r sin θ ) a θ  A / m 2 exists in a conductingmedium, where  k  is a constant. Determine the total current in the  a z  direction that crosses acircular disk of radius  R , centered on the  z  axis and located at a)  z  = 0; b)  z  =  h .Integration over a disk means that we use cylindrical coordinates. The general ﬂux integralassumes the form: I   =   s J ·  d S  =    2 π 0    R 0 − kr sin θ  a θ  · a z      − sin θ ρ d ρ d φ Then, using  r  =   ρ 2 +  z 2 , this becomes I   =    2 π 0    R 0 k ρ   ρ 2 +  z 2 = 2 π k   ρ 2 +  z 2  R 0 = 2 π k   R 2 +  z 2 − z  At  z  = 0 (part  a ), we have  I  (0) = 2 π kR  , and at  z  =  h  (part  b ):  I  ( h ) = 2 π k  √  R 2 +  h 2 − h  .60  5.7.  Assuming that there is no transformation of mass to energy or vice-versa, it is possible towrite a continuity equation for mass.a) If we use the continuity equation for charge as our model, what quantities correspond to  J and  ρ v ? These would be, respectively, mass ﬂux density in (kg / m 2 − s) and mass densityin (kg / m 3 ).b) Given a cube 1 cm on a side, experimental data show that the rates at which mass isleaving each of the six faces are 10.25, -9.85, 1.75, -2.00, -4.05, and 4.45 mg/s. If weassume that the cube is an incremental volume element, determine an approximate valuefor the time rate of change of density at its center. We may write the continuity equationfor mass as follows, also invoking the divergence theorem:   v ∂ρ m ∂  t dv  = −   v ∇ · J m  dv  = −   s J m  · dS where    s J m  · dS  = 10 . 25 − 9 . 85 + 1 . 75 − 2 . 00 − 4 . 05 + 4 . 45 = 0 . 550 mg / sTreating our 1 cm 3 volume as di ﬀ  erential, we ﬁnd ∂ρ m ∂  t. = − 0 . 550 × 10 − 3 g / s10 − 6 m 3  = − 550g / m 3 − s 5.8.  A truncated cone has a height of 16 cm. The circular faces on the top and bottom have radiiof 2mm and 0.1mm, respectively. If the material from which this solid cone is constructedhas a conductivity of 2 × 10 6 S/m, use some good approximations to determine the resistancebetween the two circular faces.Consider the cone upside down and centered on the positive  z  axis. The 1-mm radius endis at distance  z  =    from the  x - y  plane; the wide end (2-mm radius) lies at  z  =   +16cm.   is chosen such that if the cone were not truncated, its vertex would occur at the srcin.The cone surface subtends angle  θ c  from the  z  axis (in spherical coordinates). Therefore,we may write   = 0 . 1mmtan θ c and tan θ c  = 2mm160 +   Solving these, we ﬁnd    = 8 . 4 mm, tan θ c  = 1 . 19 × 10 − 2 , and so  θ c  = 0 . 68 ◦ , which givesus a very thin cone! With this understanding, we can assume that the current density isuniform with  θ  and  φ  and will vary only with spherical radius,  r . So the current densitywill be constant over a spherical cap (of constant  r ) anywhere within the cone. As thecone is thin, we can also assume constant current density over any  ﬂat   surface within thecone at a specifed  z . That is, any spherical cap looks ﬂat if the cap radius,  r , is largecompared to its radius as measured from the  z  axis ( ρ ). This is our primary assumption.Now, assuming constant current density at constant  r , and net current,  I  , we may write I   =    2 π 0    θ c 0 J  ( r ) a r  · a r  r 2 sin θ d θ d φ  = 2 π r 2 J  ( r )(1 − cos θ c )or J ( r ) =  I  2 π r 2 (1 − cos θ c )  a r  = (1 . 42 × 10 4 ) I  2 π r 2  a r 61
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