HW 07-solutions-2 - karna (pk4534) HW 07 li (59050) This...

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karna (pk4534) – HW 07 – li – (59050) 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Consider a long wire and a rectangular current loop. A B C D I 1 4 b a I 2 Determine the magnitude and direction oF the net magnetic Force exerted on the rectan- gular current loop due to the current I 1 in the long straight wire above the loop. 1. 5 F = μ 0 I 1 I 2 4 2 π ± a a + b ² ,down 2. 5 F = μ 0 I 1 I 2 2 π ± ab a + b ² 3. 5 F = μ 0 I 1 I 2 4 2 π ( a - b ), right 4. 5 F = μ 0 I 1 I 2 4 2 π ( a + b ), up 5. 5 F = μ 0 I 1 I 2 4 2 π 6. 5 F = μ 0 I 1 I 2 4 2 π ³ b a ( a + b ) ´ ,up correct 7. 5 F = μ 0 I 1 I 2 a 2 π4 ( b - a ), up 8. 5 F = μ 0 I 1 I 2 2 π ( a - b ),leFt Explanation: To compute the net Force on the loop, we need to consider the Forces on segments AB , BC , CD ,and DA .The net Force on the loop is the vector sum oF the Forces on the pieces oF the loop. The magnetic Force on AB due to the straight wire can be calculated by using 5 F AB = I 2 µ B A d5s × 5 B. In order to use this, we need to know the magnitude and direction oF the magnetic feld at each point on the wire loop. We can apply the Biot-Savart Law. The result oF this is that the magnitude oF the magnetic feld due to the straight wire is B = μ 0 I 1 2 πr , and the direction oF the magnetic feld is given by the right hand rule; the feld curls around the straight wire with the feld coming out oF the page above the wire and the feld going into the page below the wire. We can now fnd the Force on the segment AB ;app ly ing the right hand rule to fnd the direction oF the cross product, × 5 B ,weseethattheForcewill be in the up direction. Since the wire along the segment AB is straight and always at a right angle to 5 B ,thecrossproducts imp l ifes to Bds .S incethemagn itudeo Fthemagnet ic feld is constant along segment AB ,i tc an come out oF the integral which simplifes to give us the result, F AB = I 2 4B 1 = I 2 4 ± μ 0 I 1 2 πa ² . ±ollowing the same argument, we see that the Force on the segment is F = I 2 4 ³ μ 0 I 1 2 π ( a + b ) ´ , and its direction is down. This is because the direction oF the current is now in in the opposite direction along segment ! We can do the use the same procedure For segments and ,butbecausethemag- netic feld decreases with distance From the straight wire, 5 B is changing along these seg- ments. This means that the integrals are not as simple. Using the right hand rule, we see that the Force on segment is directed to- wards the right and the Force on segment is directed towards the leFt. Because the two segments oF wire are symmetrically placed, their magnitudes will be equal. Since these
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karna (pk4534) – HW 07 – li – (59050) 2 forces on the loop have equal magnitudes but opposite directions, they will cancel. Look- ing more closely at segments BC and DA , we see that for each small portion of the seg- ment ,wecanFndasma l lport iononthe segment such that the forces on these two portions are the same magnitude (The small portions are the same distance from the straight wire.); because the currents are in op-
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This note was uploaded on 11/03/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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HW 07-solutions-2 - karna (pk4534) HW 07 li (59050) This...

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