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field due to the current element is dB = 0 , where φ is the angle between 4π r 2 ! r and the current direction. EXECUTE: The magnetic field at the given points is: μ Idl sin φ μ0 (200 A)(0.000100 m) dBa = 0 = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2 dBb = dBc = dBd = dBe = μ0 Idl sin φ μ0 (200 A)(0.000100 m)sin 45° = = 0.705 × 10−6 T. 4π r 2 4π 2(0.100 m) 2 μ0 Idl sin φ μ0 (200 A)(0.000100 m) = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2 μ0 Idl sin φ μ0 Idl sin(0°) = = 0. 4π r 2 4π r2…

field due to the current element is dB = 0 , where φ is the angle between 4π r 2 ! r and the current direction. EXECUTE: The magnetic field at the given points is: μ Idl sin φ μ0 (200 A)(0.000100 m) dBa = 0 = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2 dBb = dBc = dBd = dBe = μ0 Idl sin φ μ0 (200 A)(0.000100 m)sin 45° = = 0.705 × 10−6 T. 4π r 2 4π 2(0.100 m) 2 μ0 Idl sin φ μ0 (200 A)(0.000100 m) = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2 μ0 Idl sin φ μ0 Idl sin(0°) = = 0. 4π r 2 4π r2…

field due to the current element is dB = 0 , where φ is the angle between 4π r 2 ! r and the current direction. EXECUTE: The magnetic field at the given points is: μ Idl sin φ μ0 (200 A)(0.000100 m) dBa = 0 = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2 dBb = dBc = dBd = dBe = μ0 Idl sin φ μ0 (200 A)(0.000100 m)sin 45° = = 0.705 × 10−6 T. 4π r 2 4π 2(0.100 m) 2 μ0 Idl sin φ μ0 (200 A)(0.000100 m) = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2 μ0 Idl sin φ μ0 Idl sin(0°) = = 0. 4π r 2 4π r2…

field due to the current element is dB = 0 , where φ is the angle between 4π r 2 ! r and the current direction. EXECUTE: The magnetic field at the given points is: μ Idl sin φ μ0 (200 A)(0.000100 m) dBa = 0 = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2 dBb = dBc = dBd = dBe = μ0 Idl sin φ μ0 (200 A)(0.000100 m)sin 45° = = 0.705 × 10−6 T. 4π r 2 4π 2(0.100 m) 2 μ0 Idl sin φ μ0 (200 A)(0.000100 m) = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2 μ0 Idl sin φ μ0 Idl sin(0°) = = 0. 4π r 2 4π r2…

the current element is dB = 0 , where φ is the angle between 4π r 2 ! r and the current direction. EXECUTE: The magnetic field at the given points is: μ Idl sin φ μ0 (200 A)(0.000100 m) dBa = 0 = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2 dBb = μ0 Idl sin φ μ0 (200 A)(0.000100 m)sin 45° = = 0.705 × 10−6 T. 4π r 2 4π 2(0.100 m) 2 dBc = μ0 Idl sin φ μ0 (200 A)(0.000100 m) = = 2.00 × 10−6 T. 4π r 2 4π (0.100 m) 2 dBd = μ0 Idl sin φ μ0 Idl sin(0°) = = 0.…