2 B r i π µ ⋅ = (Ley de Ampère) 2 R R r + = 2 2 R R B i π µ + ⋅ ⋅ = 2
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B ⋅ 2π r = µ0 iINT r= R1 + R2 2 B ⋅ 2π ⋅ B= (Ley de Ampère) R1 + R2 = µ0 i 2 2 µ0i 2 π ( R1 + R2 ) 4 π ⋅10 ⋅ 5 B= π ⋅ 0,13 −7 B = 1,54 ⋅10−5 [ T ] B i1 r i2 R1 R2
