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Solucionário Completo - Eletrom...para Engenheiros Com Aplicações - Ch04

Solucionário do livro "Eletromagnetismo para Engenheiros com aplicações"

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Chapter 4 Problem Solutions 4.11 . The flux in the left loop is ψ 1 = B × 01 . × 1 = 0.2t × 10 −3 Wb , and the flux in the right . × 0.5 = 01 . t × 10 −3 Wb . The induced sources are shown below: loop is ψ 2 = B × 01 dψ 1 dψ 2 V1 = = 0.2mV and V2 = = 01 . mV . Solving the resulting circuit gives dt dt 100 V = −01 . mV − 0.2mV = −0.233mV . 100 + 50 0.2 mV 0.1 mV + – + – + 50 V 100 – 4.1.2 . × 0.5 = 01 . t × 10 −3 The flux in the inner loop is ψ 1 = B × 01 Wb , and the flux in the outer loop is ψ 2 = B × 0.3 × 1 = 0.6t × 10 −3 Wb . The induced sources are shown below: dψ 1 dψ 2 V1 = = 01 . mV and V2 = = 0.6mV . Solving the resulting circuit gives dt dt 100 V = 0.6mV − 01 . mV = 0.533mV . 100 + 50 0.6 mV + – 0.1 mV + – + V 50 100 – 4-1 4.1.3 The flux in the left loop is ψ 1 = B × 0.5 × 0.2 = 10 −3 sin(2π 60t ) Wb , and the flux in the right loop is ψ 2 = B × 0.3 × 0.2 = 0.6 × 10 −3 sin(2π 60t ) Wb . The induced sources are dψ 1 = 2π 60 cos(2π × 60t ) mV and shown below: V1 = 14442444 3 dt 377 dψ 2 V2 = = 2π 60 × 0.6 cos(2π × 60t ) mV . Solving the resulting circuit gives 14444244443 dt 266 V = 226 cos(2π × 60t ) mV + 50 377 cos(2π × 60t ) mV = 3016 . cos(2π × 60t ) mV . 200 + 50 V1 V2 – + – + + 200 50 V – 4.1.4 (b) and (d) are not correct. In (b) a current must be established that will induce a secondary current I that will oppose the change in the original magnetic field. For this case the induced current must be counterclockwise to keep the field from decreasing. Similarly, in (d) a current must be established that will induce a secondary current I that will oppose the change in the original magnetic field. For this case the induced current must be clockwise to keep the field from increasing. 4.1.5 The flux in the loop is ψ = B × Area = 2 cos(2π × 60t ) × 0.5 × 10t mWb = 10t cos(2π × 60t ) mWb . The induced 4-2 voltage is V = I= dψ = 10 cos(2π × 60t ) − 3770t sin(2π × 60t ) mV . Hence the current is dt [ ] V = 01 . cos(2π × 60t ) − 37.7t sin(2π × 60t ) 100 [ ] mA . V + – I v 50 cm 100 4.1.6 t The position of the bar is L = ∫ 100 cos(10τ ) dτ = 10 sin(10t ) m . The flux in the loop is 0 ψ = B × Area = 10 × 0.5 × 10 sin(10t ) mWb = 0.05 sin(10t ) Wb . Hence the induced voltage is V = 0.5 cos(10t ) dψ = 5 cos(10t ) mA . = 0.5 cos(10t ) V . Hence the current is I = 100 dt – + I v 50 cm 100 4.1.7 A view in the xy plane is shown below. The flux through the loop is ψ = ∫ B • ds = B × Area × cos(ω t ) = 1 cos(ω t ) mWb . Hence an induced voltage in the dψ loop is V = = −ω sin(ω t ) mV = -5 sin(ω t ) mV with polarity shown below. Hence dt the current is I = V = −2.5 cos(ω t ) mA . 2 4-3 B V + – y 2 ␻t I x B 4.1.8 The magnetic flux density a distance r from an infinitely long current was obtained in the µ I (t ) and is circumferentially directed about the wire. previous chapter as B( t ) = o 2π r Hence the flux through the loop formed by the household power wiring is 3m 1.09km µ I ( t ) . km   109 o . × 10 −8 I ( t ) . ψ = ∫ B • ds = ∫ drdz = 6 × 10 −7 ln   I ( t ) = 517 ∫   1km z = 0 r=1km 2π r The induced voltage is V = dψ dI ( t ) = 517 . × 10 −8 and is sketched below. Hence dt dt V = 2,585V for 0 < t < 1µ s and is V = −287V for 1µ s < t < 10µ s . V (t) 2585 V 10 ␮s –287 V t 1 ␮s 4.1.9 The magnetic flux density threading the loop is B = µoI . Assuming that the loop starts 2π r at t=0 barely touching the wire, at some time t the flux through the loop is (downward) 4-4 l vt + w µ I o drdz = µ o Il ln  vt + w    ∫  vt  2π z = 0 r = vt 2π r ψ = ∫ B • ds = ∫ . The induced voltage in the loop (tending to push current counter clockwise) is µ o Ilw dψ d µ o Il  vt + w  µ o Il  vt   w  V= = ln  . Hence =  − 2  = −  vt  2π  vt + w   vt  2π ( vt + w)t dt dt 2π I=− µ o Ilw V . = R 2π R( vt + w)t 4.110 . A sketch of the problem looking down on the plane of rotation is shown below. The further we go out along the bar length, the greater the velocity cutting the magnetic field lines. The linear velocity of a section of the bar at a radius r is v = ω r . The voltage l induced in the bar is ∫ ( v × B ) • dl = ∫ Bω rdr = r =0 Bω l 2 . By the right-hand rule, the 2 rotating end is the positive end. ␻ l B 4.111 .  ∂ Ez ∂ E y   ∂ E y ∂ Ex   ∂ Ex ∂ Ez  Forming ∇ × E =  − − − ax +   a . From the a y +  ∂z  ∂x  ∂y  z  ∂z  ∂y  ∂x form of E this reduces to ∂ Ey ∂ Ey ∇×E= − ax + a = β E m sin α x cos(ω t − β z )a x + α E m cos α x sin(ω t − β z )a z ∂z ∂x z 4-5 ∂H we obtain ∂t E β E α H = − m sin α x sin(ω t − β z )a x + m cos α x cos(ω t − β z )a z . ω µo ω µo . From Faraday’s law: ∇ × E = − µ o 4.112 .  ∂ Ez ∂ E y   ∂ E y ∂ Ex   ∂ Ex ∂ Ez  Forming ∇ × E =  − − − ax +   a . From the a y +  ∂z  ∂x  ∂y  z  ∂z  ∂y  ∂x ∂ Ex a = β E m cos β z cos(ω t )a y . From Faraday’s form of E this reduces to ∇ × E = ∂z y ∂H E β law: ∇ × E = − µ o we obtain H = − m cos β z sin(ω t )a y . ∂t ω µo 4.21 . Above a frequency for which σ = 1 the displacement current dominates the ω ε rε o conduction current. This occurs for f > 45kHz . 4.2.2  ∂ Hz ∂ H y   ∂ H y ∂ Hx   ∂ Hx ∂ Hz  Forming ∇ × H =  − − − ax +   a . From a y +  ∂z  ∂x  ∂y  z  ∂z  ∂y  ∂x the form of H this reduces to  ∂ Hx ∂ Hz  ∇×H = −  a = ( − β H x + α H z ) sin α x cos(ω t − β z )a y . From Ampere’s ∂x  y  ∂z law: ∇ × H = ε o (− β H x + α H z ) sin α x sin ω t − β z a ∂E we obtain E = ( ) y ωε o ∂t 4.2.3  ∂ Hz ∂ H y   ∂ H y ∂ Hx   ∂ Hx ∂ Hz  Forming ∇ × H =  − − − ax +   a . From a y +  ∂z  ∂x  ∂y  z  ∂z  ∂y  ∂x ∂ Hy a = β H m sin β z sin(ω t )a x . From the form of H this reduces to ∇ × H = − ∂z x ∂E β Hm Ampere’s law: ∇ × H = ε o sin β z cos(ω t )a x we obtain E = − ∂t ωε o 4-6 4.31 . Gauss’ law becomes, in rectangular coordinates, ∇ • (ε o E) = 0 = ∂ Ex ∂ E y ∂ Ez + + . ∂y ∂z ∂x But for the given electric field, E = Em sin α x sin(ω t − β z )a y , it has only a y component which is independent of y. Hence, the divergence is zero. 4.3.2 Gauss’ law becomes, in rectangular coordinates, ∇ • (ε o E) = 0 = ∂ Ex ∂ E y ∂ Ez + + . ∂y ∂z ∂x But for the given electric field, E = Em sin β z cos(ω t )a x , it has only an x component which is independent of x. Hence, the divergence is zero. 4.3.3 Gauss’ law becomes, in rectangular coordinates, ∇ • ( µ o H ) = 0 = ∂ Hx ∂ H y ∂ Hz + + . ∂y ∂z ∂x For the given magnetic field, H = H x sin α x sin(ω t − β z )a x + H z cos α x cos(ω t − β z )a z , so that ∂ Hx ∂ Hz + = α H x cos α x sin(ω t − β z )a x + β H z cos α x sin(ω t − β z )a z . Hence the ∂x ∂z divergence is zero and Gauss’ law is satisfied only if α H x + β H y = 0 . 4.3.4 Gauss’ law becomes, in rectangular coordinates, ∇ • ( µ o H ) = 0 = ∂ Hx ∂ H y ∂ Hz + + . ∂y ∂z ∂x But for the given magnetic field, H = H m cos β z sin(ω t )a y , the y component is independent of y. Hence, the divergence is zero. 4.61 . π  1 π  W 1  . e −8 z  cos 2ω t − 8 z −  + cos  a z 2 . The Poynting vector is S = E × H = 715   4 m 4 2 2 715 . −8 z W π The Poynting vector averaged over one cycle is S average = e cos  a z 2 . This  4 2 m is perpendicular only to the sides at z=0 and z=1m. Hence the average power exiting the cube is 4-7 Paverage = −Saverage =- z =0 × Area + Saverage z =1m × Area 715 . −8( z =0)  π  715 . −8( z =1m)  π  e e cos  + cos  = -25.27W  4 2  4 2 4.71 . The components of E that are tangential to the boundary must be continuous. Hence E 2, tan = β a y + γ a z . The components of D that are normal to the boundary must be continuous. Hence D1, norm = ε 1E1, norm = ε 1α a x = D 2, norm = ε 2 E 2, norm so that ε ε E 2, norm = 1 α a x . Therefore, E 2 x = 0 = E 2, norm + E 2, tan = 1 α a x + β a y + γ a z . ε2 ε2 4.7.2 The components of B that are normal to the boundary must be continuous. Hence B 2, norm = α a x . The components of H that are tangential to the boundary must be 1 β γ 1 B1, tan = ay + a z = H 2, tan = B 2, tan so that continuous. Hence H1, tan = µ1 µ1 µ1 µ2 µ β a y + 2 γ a z . Therefore, µ1 µ1 µ µ = B 2, norm + B 2, tan = α a x + 2 β a y + 2 γ a z . µ1 µ1 µ2 B 2, tan = B2 x =0 4.7.3 The components of D that are normal to the boundary must be continuous. Hence D 2, norm = α a x . The components of E that are tangential to the boundary must be 1 1 1 1 D1, tan = D 2, tan so that β a y + γ a z = E 2, tan = continuous. Hence E1, tan = ε1 ε1 ε ε D 2, tan = 2 β a y + 2 γ a z . Therefore, ε1 ε1 ε2 ε1 ε ε D 2 x = 0 = D 2,tan + D 2,norm = α a x + β 2 a y + γ 2 a z . ε ε 1 1 4.7.4 The components of H that are tangential to the boundary must be continuous. Hence H 2, tan = β a y + γ a z . The components of B that are normal to the boundary must be continuous. Hence B1, norm = µ1H1, norm = µ1α a x = B 2, norm = µ 2 H 2, norm so that 4-8 µ2 µ β a y + 2 γ a z . Therefore, µ1 µ1 µ µ = B 2, norm + B 2, tan = α a x + 2 β a y + 2 γ a z . µ1 µ1 B 2, tan = B2 x =0 4.7.5 The tangential components of E must be zero at the surface of a perfect conductor. Hence γ = 0 . Also, the x and y components of E must be equal in order that there be no tangential component from these two components and hence α = β . the normal components of B must be zero at the surface of a perfect conductor. Hence the x and y components must form a resultant that is tangent to the surface so that σ = −δ . 4.81 . The components tangent to the plane are 2a x − 3a y . Reversing these gives −2a x + 3a y . Similarly reversing the z component gives −4a z . Hence the image current is I image = −2a x + 3a y − 4a z A at (0,0,-2). 4.91 . 6 To convert a phasor quantity to the time domain we simply multiply by e jω t = e j 2π ×10 t and take the real part of the result. Hence, (a) ( ) ( ) $ e jω t = Re − j30e jω t a + Re − 10 e jω t a  E = Re E x y  j  π π   = −30 cos 2π × 10 6 t +  a x + 10 cos 2π × 10 6 t +  a y   2 2 , (b) ( ) ( ) 4π $ e jω t = Re10e jω t e j 3  a = 10 cos 2π × 10 6 t + 4π  a , H = Re H z z  3   (c) ( ) ( ) B = Re B$ e jω t = Re 4e −2 z e jω t e − j 4πz a x = 4e −2 z cos 2π × 10 6 t − 4π z a x , $ = 10e −3x e − j 3π z a , (e) B$ = 5 sin(3z )e − j 6 z a . (d) E y x 4-9 4.9.2 $ = E sin α xe − jβ z a . Faraday’s law in phasor form is The phasor form of the field is E m y $ = − jω µ H $ ∇×E o . Writing out the curl gives  ∂ E$  ∂ E$ y ∂ E$  ∂ E$ y   ∂ E$ x ∂ E$ z  z x $    ∇×E= −  a x +  ∂ z − ∂ x  a y +  ∂ x − ∂ y  a z . From the form of ∂ y ∂ z     the field this reduces to ∂ E$ y ∂ E$ y $ =− ∇×E ax + a z = jβ Em sin α xe − jβ z a x + α Em cos α xe − jβ z a z . Hence ∂z ∂x $ = − 1 ∇×E $ = j 1 ∇×E $ = j 1 jβ E m sin α xe − jβ z a x + α E m cos α xe − jβ z a z H jω µ o ω µo ω µo β Em α Em =− sin α xe − jβ z a x + j cos α xe − jβ z a z ( ω µo ω µo The time-domain field is $ e jω t H = Re H ( ) β Em α Em π  sin α x cos(ω t − β z )a x + cos α x cos ω t − β z +  a z  2 ω µo ω µo β Em α Em =− sin α x cos(ω t − β z )a x − cos α x sin(ω t − β z )a z ω µo ω µo =− 4.9.3 ( ) ( ) $ jω t and H = Re He $ jω t . The instantaneous Consider two field vectors, E = Re Ee power density or Poynting vector is S = E × H . Rewriting the field vectors as $ jω t = 1 Ee $ jω t + E $ ∗e − jω t and substituting into S = E × H yields E = Re Ee 2 1 $ $ ∗ $∗ $ 1 $ $ j 2ω t $ ∗ $ ∗ − j 2ω t S = E×H = E×H +E ×H + E × He +E ×H e . But since 4 4 ( ) ( ( ) ( ) ) (E$ × H$ ∗ ) = (E$ ∗ × H$ ) and A$ + A$ ∗ = 2 Re(A$ ) , this may be written as T 1 $ ×H $ ∗ + 1 Re E $ ×H $ e j 2ω t . The time average of this is S = 1 ∫ S dt and T S = Re(E ) 2 ( ) av T0 2 ∗ is the period of the sinusoid. The first term of our result is a constant and the second term is sinusoidal and hence averages to zero giving the desired result. 4-10 ) 4.10.1 The magnetic flux density in the core due to a current I is, from Chapter 3, Bφ = µrµoI 2π r where the flux is circumferentially directed and r is approximately the mean radius of the core. The total flux through the windings is approximately ψ = BA where A is the area of the core cross section. According to Faraday’s law a voltage will be generated in the turns dψ µ r µ o NA dI dI V$ ω µ r µ o NA of wire of V = N . But = . ⇔ jω I$ . Hence Z$T = = 2π r dt 2π r dt dt I$ Substituting numerical values gives Z$T = (ω = 2π f )( µ r ( = 200)µ o ( N = 10) A = 4 × 10 −4 2π ( r = 0.02) 4-11 ) = 50.3Ω = 34dBΩ