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Unformatted text preview: dang (rtd356) – HW 07 – li – (59050) 1 This printout should have 32 questions. Multiplechoice 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 ℓ 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. vector F = μ I 1 I 2 ℓ 2 π parenleftbigg a a + b parenrightbigg , down 2. vector F = μ I 1 I 2 ℓ 2 π ( a + b ), up 3. vector F = μ I 1 I 2 ℓ 2 π ( a − b ), right 4. vector F = μ I 1 I 2 ℓ 2 π a b , down 5. vector F = μ I 1 I 2 2 π parenleftbigg a b a + b parenrightbigg , down 6. vector F = μ I 1 I 2 2 π ( a − b ),left 7. vector F = μ I 1 I 2 ℓ 2 π bracketleftbigg b a ( a + b ) bracketrightbigg , up correct 8. vector F = μ I 1 I 2 a 2 π ℓ ( b − a ), up 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 vector F AB = I 2 integraldisplay B A dvectors × vector 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 BiotSavart Law. The result of this is that the magnitude of the magnetic field due to the straight wire is B = μ 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, dvectors × vector 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 vector 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 ℓ B 1 = I 2 ℓ parenleftbigg μ I 1 2 π a parenrightbigg . Following the same argument, we see that the force on the segment CD is F CD = I 2 ℓ bracketleftbigg μ I 1 2 π ( a + b ) bracketrightbigg , 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, vector 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,...
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 Fall '08
 Turner
 Magnetic Field

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