ohw18-solutions

# ohw18-solutions - Reilly(mar3978 ohw18 turner(56725 This...

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Reilly (mar3978) – ohw18 – turner – (56725) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A closed circuit consists of two semicircles of radii 80 cm and 41 cm that are connected by straight segments, as shown in the figure. A current of I = 4 A flows around this circuit in the clockwise direction. The permeability of free space is 4 π × 10 7 T m / A. O 41 cm 80 cm ˆ k ˆ ı ˆ ˆ is upward from the paper I = 4 A I What is the direction of the magnetic field B at point O . 1. vector B bardbl vector B bardbl = +ˆ 2. vector B bardbl vector B bardbl = +ˆ ı 3. vector B bardbl vector B bardbl = ˆ k 4. None of these are true‘. 5. vector B bardbl vector B bardbl = ˆ correct 6. vector B bardbl vector B bardbl = ˆ ı 7. vector B bardbl vector B bardbl = + ˆ k Explanation: If you curve the fingers on your right hand in the direction of the current, the magnetic field points along your thumb. 002 (part 2 of 2) 10.0 points Find the magnitude of the magnetic field at point at point O . Correct answer: 4 . 63576 μ T. Explanation: Let : r 1 = 80 cm = 0 . 8 m , r 2 = 41 cm = 0 . 41 m , I = 4 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  , and the magnetic field due to the arc 2 is vector B 2 = μ 0 I 4 r 2 ˆ  . 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 = (4 π × 10 7 T m / A) (4 A) 4 × parenleftbigg 1 0 . 8 m + 1 0 . 41 m parenrightbigg ˆ parenleftbigg 10 6 μ T 1 T parenrightbigg = ( 4 . 63576 μ T) ˆ bardbl vector B bardbl = 1 . 003 10.0 points The closed loop shown in the figure carries a current of 14 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.

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Reilly (mar3978) – ohw18 – turner – (56725) 2 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 . 22173 × 10 6 T . Explanation: Let : r i = 40 cm = 0 . 4 m , r o = 60 cm = 0 . 6 m , I = 14 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) (14 A) 12 × parenleftbigg 1 0 . 6 m 1 0 . 4 m parenrightbigg ˆ ı = ( 1 . 22173 × 10 6 T ) ˆ ı .
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