Example An infinite filament on the z axis carries 207 mA in the a, direction. Three uniform cylindrical current sheets are also present: 400 mA/m at p= 1 cm, -250 mA/m at p = 2 cm, and -300 mA/m at p = 3 cm. Calculate Ho at p = 0.5, 1.5, 2.5, and 3.5 cm: We find H, at each of the required radii by applying Ampere's circuital law to circular paths of those radii; the paths are centered on the z axis. So, at p₁ = 0.5 cm: fH. dL = 27P1H61 = Lenci = 20 × 10-³ A Thus 10 x 10-³ Hol P1 At p = p2 = 1.5 cm, we enclose the first of the current cylinders at p = 1 cm. Ampere's law becomes: Following this method, at 2.5 cm: 272 Ho2 = 207 +27(10-2) (400) mA = H62 = and at 3.5 cm, H63 = H64 10 x 10-3 0.5 x 10-2 10+4.00 (2 x 10-²)(250) 2.5 x 10-2 = = 2.0 A/m 10+ 4.00 1.5 x 10-2 = 360 mA/m 10+4.00-5.00 - (3 × 10-²) (300) 3.5 x 10-2 = 933 mA/m = 0

Example An infinite filament on the z axis carries 207 mA in the a, direction. Three uniform cylindrical current sheets are also present: 400 mA/m at p= 1 cm, -250 mA/m at p = 2 cm, and -300 mA/m at p = 3 cm. Calculate Ho at p = 0.5, 1.5, 2.5, and 3.5 cm: We find H, at each of the required radii by applying Ampere's circuital law to circular paths of those radii; the paths are centered on the z axis. So, at p₁ = 0.5 cm: fH. dL = 27P1H61 = Lenci = 20 × 10-³ A Thus 10 x 10-³ Hol P1 At p = p2 = 1.5 cm, we enclose the first of the current cylinders at p = 1 cm. Ampere's law becomes: Following this method, at 2.5 cm: 272 Ho2 = 207 +27(10-2) (400) mA = H62 = and at 3.5 cm, H63 = H64 10 x 10-3 0.5 x 10-2 10+4.00 (2 x 10-²)(250) 2.5 x 10-2 = = 2.0 A/m 10+ 4.00 1.5 x 10-2 = 360 mA/m 10+4.00-5.00 - (3 × 10-²) (300) 3.5 x 10-2 = 933 mA/m = 0