OHW 21 - terry(ect328 – oldhomework 21 – Turner...

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Unformatted text preview: terry (ect328) – oldhomework 21 – Turner – (59130) 1 This print-out should have 13 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 75 cm and 37 cm that are connected by straight segments, as shown in the figure. A current of I = 5 A flows around this circuit in the clockwise direction. The permeability of free space is 4 π × 10 − 7 T m / A. O 3 7 c m 7 5 c m ˆ ı ˆ ˆ k ˆ k is upward from the paper I = 5 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 correct 4. vector B bardbl vector B bardbl = + ˆ k 5. vector B bardbl vector B bardbl = +ˆ ı 6. vector B bardbl vector B bardbl = +ˆ 7. None of these are true‘. 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: 6 . 33979 μ T. Explanation: Let : r 1 = 75 cm = 0 . 75 m , r 2 = 37 cm = 0 . 37 m , I = 5 A , and μ = 4 π × 10 − 7 T m / A . The magnetic field due to a circular loop at its center is B = μ 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 I 4 r 1 ˆ k , and the magnetic field due to the arc 2 is vector B 2 = − μ I 4 r 2 ˆ k . Thus the total magnetic field is vector B = vector B 1 + vector B 2 = − μ I 4 parenleftbigg 1 r 1 + 1 r 2 parenrightbigg ˆ k = − (4 π × 10 − 7 T m / A) (5 A) 4 × parenleftbigg 1 . 75 m + 1 . 37 m parenrightbigg ˆ k parenleftbigg 10 6 μ T 1 T parenrightbigg = ( − 6 . 33979 μ T) ˆ k bardbl vector B bardbl = 5 . 003 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 5 7 π O r terry (ect328) – oldhomework 21 – Turner – (59130) 2 What is the magnitude of the magnetic field B O at the origin O due to the current through this path? 1. B O = 3 20 μ I r 2. B O = 7 48 μ I r 3. B O = 1 5 μ I r 4. B O = 0 5. B O = 1 6 μ I r 6. B O = 3 16 μ I r 7. B O = 1 7 μ I r 8. B O = 5 24 μ I r 9. B O = 5 28 μ I r correct 10. B O = 7 52 μ I r Explanation: Using the Biot-Savart law, the magnetic field due to the two radial legs is zero since dvectors × ˆ r = 0 ....
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This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas.

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OHW 21 - terry(ect328 – oldhomework 21 – Turner...

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