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Example A long straight non-magnetic conductor of 0.2 mm radius carries a uniformly-distributed current of 2 A dc. a) Find J within the conductor: Assuming the current is +z directed, = 1.59 x 10¹ a. A/m² V x H = J= b) Use Ampere's circuital law to find H and B within the conductor: Inside, at radius p, we have ΣπρΗ = πρ1 = H H = º/ª = 7.96 × 10%µ a¿ A/m Then B = H = (47 x 10-7)(7.96 x 106)pa = 10pa, Wb/m². c) Show that V x H = J within the conductor: Using the result of part b, we find, 1 d pdp 2 (0.2 x 10-3)2 (pH₂) az 1 d (1.59 x 107p² 2 ; (³² p dp == H= 0²p²); Now BoH = μ0/(лp) as Wb/m². d) Find H and B outside the conductor (note typo in book): Outside, the entire current is enclosed by a closed path at radius p, and so Σπρ az 1.59 x 107 a A/m² = J 1 a = a A/m др

Example A long straight non-magnetic conductor of 0.2 mm radius carries a uniformly-distributed current of 2 A dc. a) Find J within the conductor: Assuming the current is +z directed, = 1.59 x 10¹ a. A/m² V x H = J= b) Use Ampere's circuital law to find H and B within the conductor: Inside, at radius p, we have ΣπρΗ = πρ1 = H H = º/ª = 7.96 × 10%µ a¿ A/m Then B = H = (47 x 10-7)(7.96 x 106)pa = 10pa, Wb/m². c) Show that V x H = J within the conductor: Using the result of part b, we find, 1 d pdp 2 (0.2 x 10-3)2 (pH₂) az 1 d (1.59 x 107p² 2 ; (³² p dp == H= 0²p²); Now BoH = μ0/(лp) as Wb/m². d) Find H and B outside the conductor (note typo in book): Outside, the entire current is enclosed by a closed path at radius p, and so Σπρ az 1.59 x 107 a A/m² = J 1 a = a A/m др

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Example A long straight non-magnetic conductor of 0.2 mm radius carries a uniformly-distributed current of 2 A dc. a) Find J within the conductor: Assuming the current is +z directed, = 1.59 × 107 a- A/m² J= b) Use Ampere's circuital law to find H and B within the conductor: Inside, at radius p, we have ΣπρΗ = πρ1 = H H = º/ª = 7.96 × 10%µ a¿ A/m V x H = 2 (0.2 x 10-3)2 Then B = oH = (47 x 10-7)(7.96 x 106)pa = 10pa, Wb/m². c) Show that V x H = J within the conductor: Using the result of part b, we find, 1 d pdp pap (² 1 d (1.59 x 107p² 2 (pH₂) az == H= · 10²0²), Now BoH = μ0/(лp) as Wb/m². d) Find H and B outside the conductor (note typo in book): Outside, the entire current is enclosed by a closed path at radius p, and so Σπρ az 1.59 x 107 a A/m² = J 1 a = a A/m др
Transcribed Image Text:Example A long straight non-magnetic conductor of 0.2 mm radius carries a uniformly-distributed current of 2 A dc. a) Find J within the conductor: Assuming the current is +z directed, = 1.59 × 107 a- A/m² J= b) Use Ampere's circuital law to find H and B within the conductor: Inside, at radius p, we have ΣπρΗ = πρ1 = H H = º/ª = 7.96 × 10%µ a¿ A/m V x H = 2 (0.2 x 10-3)2 Then B = oH = (47 x 10-7)(7.96 x 106)pa = 10pa, Wb/m². c) Show that V x H = J within the conductor: Using the result of part b, we find, 1 d pdp pap (² 1 d (1.59 x 107p² 2 (pH₂) az == H= · 10²0²), Now BoH = μ0/(лp) as Wb/m². d) Find H and B outside the conductor (note typo in book): Outside, the entire current is enclosed by a closed path at radius p, and so Σπρ az 1.59 x 107 a A/m² = J 1 a = a A/m др
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