# Correct answer 3 5591 μ t explanation let r 1 51 cm

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Correct answer: 3 . 5591 μ T. Explanation: Let : r 1 = 51 cm = 0 . 51 m , r 2 = 27 cm = 0 . 27 m , I = 2 A , and μ 0 = 4 π × 10 7 T m / A . The magnetic field due to a circular loop at its center is B = μ 0 I 2 R . Because the loop is only half of a current loop, the magnetic field due to the arc 1 is vector B 1 = μ 0 I 4 r 1 ˆ k , and the magnetic field due to the arc 2 is vector B 2 = μ 0 I 4 r 2 ˆ k . Thus the total magnetic field is vector B = vector B 1 + vector B 2 = μ 0 I 4 parenleftbigg 1 r 1 + 1 r 2 parenrightbigg ˆ k = (4 π × 10 7 T m / A) (2 A) 4 × parenleftbigg 1 0 . 51 m + 1 0 . 27 m parenrightbigg ˆ k parenleftbigg 10 6 μ T 1 T parenrightbigg = ( 3 . 5591 μ T) ˆ k bardbl vector B bardbl = 4 . 007 10.0 points The closed loop shown in the figure carries a current of 16 A in the counterclockwise direc- tion. The radius of the outer arc is 60 cm , that of the inner arc is 40 cm . Both arcs ex- tent an angle of 60 . The permeability of free space is 4 π × 10 7 T m / A.
karna (pk4534) – HW 07 – li – (59050) 5 40 cm C B 60 cm D A O 60 y z Find the component of the magnetic field at point O along the x axis. Correct answer: 1 . 39626 × 10 6 T . Explanation: Let : r i = 40 cm = 0 . 4 m , r o = 60 cm = 0 . 6 m , I = 16 A , and μ 0 = 4 π × 10 7 T m / A . The magnetic field due to a circular loop at its center is B = μ 0 I 2 R . Because the loop is only one-sixth of a current loop, the magnetic field due to the inner arc is vector B i = μ 0 I 12 r i ˆ ı , and the magnetic field due to the outter arc is vector B o = μ 0 I 12 r o ˆ ı . Thus the total magnetic field is vector B = vector B i + vector B o = μ 0 I 12 parenleftbigg 1 r o 1 r i parenrightbigg ˆ ı = (4 π × 10 7 T m / A) (16 A) 12 × parenleftbigg 1 0 . 6 m 1 0 . 4 m parenrightbigg ˆ ı = ( 1 . 39626 × 10 6 T ) ˆ ı . 008 10.0 points Consider two radial legs extending to infinity and a circular arc carrying a current I as shown below. x y I I I I 6 11 π O r What is the magnitude of the magnetic field B O at the origin O due to the current through this path? 1. B O = 5 32 μ 0 I r 2. B O = 0 3. B O = 3 22 μ 0 I r correct 4. B O = 5 36 μ 0 I r 5. B O = 1 7 μ 0 I r 6. B O = 3 20 μ 0 I r 7. B O = 3 14 μ 0 I r 8. B O = 1 5 μ 0 I r 9. B O = 5 28 μ 0 I r 10. B O = 5 24 μ 0 I r Explanation: Using the Biot-Savart law, the magnetic field due to the two radial legs is zero since dvectors × ˆ r = 0 . However, around the arc dvectors × ˆ r = dvectors = r d vector θ . The magnetic field at at the center of an arc with a current I is B = μ 0 I 4 π integraldisplay dvectors × ˆ r r 2 = μ 0 I 4 π r 2 integraldisplay ds
karna (pk4534) – HW 07 – li – (59050) 6 = μ 0 I 4 π r 2 integraldisplay r dθ = μ 0 I 4 π r integraldisplay 6 11 π 0 = μ 0 I 4 π r θ vextendsingle vextendsingle vextendsingle vextendsingle 6 11 π 0 = μ 0 I 4 π r parenleftbigg 6 11 π 0 parenrightbigg = 3 22 μ 0 I r . 009 10.0 points Three very long wires are strung parallel to each other as shown in the figure below. Each wire is at a distance r from the other two, and each wire carries a current of magnitude I in the directions shown in the figure.

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