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Q6: When x, y and z are positive and less than 6, a certain magnetic field intensity may be expressed as H = 6x(1+y)'z ax + 3x²(1+z)³ ay - (x+1)yz² az. Find the total current that crosses the strip, x = 3, 1 ≤ y ≤ 5, 3 ≤ z≤5, by using Stokes' theorem.

Q6: When x, y and z are positive and less than 6, a certain magnetic field intensity may be expressed as H = 6x(1+y)'z ax + 3x²(1+z)³ ay - (x+1)yz² az. Find the total current that crosses the strip, x = 3, 1 ≤ y ≤ 5, 3 ≤ z≤5, by using Stokes' theorem.

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"I don't want the solution using curl".I want the solution using the other side of the equation."

Q6: When x, y and z are positive and less than 6, a certain magnetic field intensity may be expressed as H = 6x(1+y)'z ax + 3x²(1+z)³ ay - (x+1)yz² az. Find the total current that crosses the strip, x = 3,1 ≤ y ≤ 5, 3 ≤ z≤5, by using Stokes' theorem.
Transcribed Image Text:Q6: When x, y and z are positive and less than 6, a certain magnetic field intensity may be expressed as H = 6x(1+y)'z ax + 3x²(1+z)³ ay - (x+1)yz² az. Find the total current that crosses the strip, x = 3,1 ≤ y ≤ 5, 3 ≤ z≤5, by using Stokes' theorem.
Step 1: given information Given data, H = 6x(1+y)¹zāx+3x²(1+z)³ā,-(x+1)yz²a₂ strip:x=3,1 ≤ y ≤ 5, 3 ≤z≤5. Step 2: stokes theorem According to Ampere law, fH.d = 1, ds = dydz .......... Now, using the stokes theorem, ff(H).ds.. |= curt of H VxH = ax d O Ə dx dy dz Ax Ay A y = az ax д əx 6x(1+y)4z 3x²(1+z)³+1)yz² ..(ii) d ду = [+1)2²-9x²(1+z)²]a, y=1 z 3 "I don't want this solution, but I want the same outcome." Step 3: total current valu only a, term we need, the remaining term a and â will be zero after taking dot of âx. O -(- +1)yz2 - - 02³x²(1+2)³73x dy (12775) =-16937.33 A From equations (i), (ii), the remaining term â, and âill be zero after taking dot of â | = = √³ ³ [-(x+1)2²—9x²(1+z)²]³. dy.dz a, = y 1 z 3 - (42²+81(1+z)²dy.dz (42²+81(1+z)²dz =-5dy y = 1 -45 (4z²+81(1+z)²dz z=3 42³ 81(1+z)³15 - -4 43² +. + 25³ 1. 3 Jz=3 az ə Əz .(iii)
Transcribed Image Text:Step 1: given information Given data, H = 6x(1+y)¹zāx+3x²(1+z)³ā,-(x+1)yz²a₂ strip:x=3,1 ≤ y ≤ 5, 3 ≤z≤5. Step 2: stokes theorem According to Ampere law, fH.d = 1, ds = dydz .......... Now, using the stokes theorem, ff(H).ds.. |= curt of H VxH = ax d O Ə dx dy dz Ax Ay A y = az ax д əx 6x(1+y)4z 3x²(1+z)³+1)yz² ..(ii) d ду = [+1)2²-9x²(1+z)²]a, y=1 z 3 "I don't want this solution, but I want the same outcome." Step 3: total current valu only a, term we need, the remaining term a and â will be zero after taking dot of âx. O -(- +1)yz2 - - 02³x²(1+2)³73x dy (12775) =-16937.33 A From equations (i), (ii), the remaining term â, and âill be zero after taking dot of â | = = √³ ³ [-(x+1)2²—9x²(1+z)²]³. dy.dz a, = y 1 z 3 - (42²+81(1+z)²dy.dz (42²+81(1+z)²dz =-5dy y = 1 -45 (4z²+81(1+z)²dz z=3 42³ 81(1+z)³15 - -4 43² +. + 25³ 1. 3 Jz=3 az ə Əz .(iii)
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