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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 – find all choices before answering. 001 10.0 points Consider a long wire and a rectangular current loop. A B C D I 1 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. F = μ 0 I 1 I 2 2 π a a + b , down 2. F = μ 0 I 1 I 2 2 π a b a + b , down 3. F = μ 0 I 1 I 2 2 π ( a - b ), right 4. F = μ 0 I 1 I 2 2 π ( a + b ), up 5. F = μ 0 I 1 I 2 2 π a b , down 6. F = μ 0 I 1 I 2 2 π b a ( a + b ) , up correct 7. F = μ 0 I 1 I 2 a 2 π ( b - a ), up 8. 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 F AB = I 2 B A d s × B . In order to use this, we need to know the magnitude and direction of the magnetic field 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 field due to the straight wire is B = μ 0 I 1 2 π r , and the direction of the magnetic field is given by the right hand rule; the field curls around the straight wire with the field coming out of the page above the wire and the field going into the page below the wire. We can now find the force on the segment AB ; applying the right hand rule to find the direction of the cross product, d s × B , we see that the force will be in the up direction. Since the wire along the segment AB is straight and always at a right angle to B , the cross product simplifies to B ds . Since the magnitude of the magnetic field is constant along segment AB , it can come out of the integral which simplifies to give us the result, F AB = I 2 B 1 = I 2 μ 0 I 1 2 π a . Following the same argument, we see that the force on the segment CD is F CD = I 2 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 CD ! We can do the use the same procedure for segments BC and DA , but because the mag- netic field decreases with distance from the straight wire, 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 BC is directed to- wards the right and the force on segment DA 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 BC , we can find a small portion on the segment DA 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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