OHW 25 - terry (ect328) – oldhomework 25 – Turner –...

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Unformatted text preview: terry (ect328) – oldhomework 25 – 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 rectangular loop coplanar to and located a distance 3 . 46 cm from a long wire carrying a current of 0 . 0905 A is shown in the figure. The wire is parallel to the longest side of the loop. 3 . 46cm 7 . 38 cm 30 . 8cm . 0905A Find the total magnetic flux through the loop. Correct answer: 6 . 36628 × 10 − 9 Wb. Explanation: Let : c = 3 . 46 cm , a = 7 . 38 cm , b = 30 . 8 cm , and I = 0 . 0905 A . c a b r dr I From Amp` ere’s law, the strength of the magnetic field created by the current-carrying wire at a distance r from the wire is B = μ I 2 π r , so the field varies over the loop and is directed perpendicular to the page. Since vector B is parallel to d vector A , the magnetic flux through an area element dA is Φ ≡ integraldisplay B dA = integraldisplay μ I 2 π r dA . Note: vector B is not uniform (it depends on r ), so it cannot be removed from the integral. In order to integrate, the area element shaded in the figure as dA = b dr . Since r is the only variable that now appears in the integral, the magnetic flux is Φ B = μ I 2 π b integraldisplay a + c c d r r = μ I b 2 π ln r vextendsingle vextendsingle vextendsingle a + c c = μ I b 2 π ln parenleftbigg a + c c parenrightbigg = μ (0 . 0905 A)(0 . 308 m) 2 π ln parenleftbigg a + c c parenrightbigg = μ (0 . 0905 A)(0 . 308 m) 2 π (1 . 14197) = 6 . 36628 × 10 − 9 Wb . 002 (part 2 of 2) 10.0 points What is the direction of the magnetic field through the rectangular loop? 1. into the plane of the paper correct 2. Parallel to the plane of the paper and parallel to the current direction in the wire. 3. out of the plane of the paper 4. Parallel to the plane of the paper and perpendicular to the current direction in the wire. Explanation: terry (ect328) – oldhomework 25 – Turner – (59130) 2 Using the right-hand rule on the long straight wire, if the current flows upward (downward), the magnetic field would be into (out of) the plane of the paper. Consequently the magnetic field is directed into the plane of the paper. 003 (part 1 of 4) 10.0 points The circular loop of wire shown in the figure is placed in a spatially uniform magnetic field such that the plane of the circular loop is per- pendicular to the direction for the magnetic field as shown in the figure. The magnetic field vector B ( t ) varies with time, with the time de- pendence given by B ( t ) = a + b t , where a = 0 . 17 T and b = 0 . 028 T / s. The acceleration due to gravity is 9 . 8 m / s 2 ....
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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 25 - terry (ect328) – oldhomework 25 – Turner –...

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