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magnetics hw3333

# magnetics hw3333 - ragsdale(zdr82 HW7 ditmire(58335 This...

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ragsdale (zdr82) – HW7 – ditmire – (58335) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A wire carrying a current 30 A has a length 0 . 1 m between the pole faces of a magnet at an angle 60 (see the figure). The magnetic field is approximately uniform at 0 . 5 T. We ignore the field beyond the pole pieces. θ I B What is the force on the wire? Correct answer: 1 . 29904 N. Explanation: Let : I = 30 A , = 0 . 1 m , θ = 60 , and B = 0 . 5 T . we use F = I ℓ B sin θ , so F = I ℓ B sin θ = (30 A) (0 . 1 m) (0 . 5 T) sin 60 = 1 . 29904 N . 002 10.0 points The magnetic force on a straight 0.45 m seg- ment of wire carrying a current of 4.5 A is 0.31 N. What is the magnitude of the component of the magnetic field that is perpendicular to the wire? Correct answer: 0 . 153086 T. Explanation: Let : = 0 . 45 m , I = 4 . 5 A , and F m = 0 . 31 N . The magnetic force is F m = I ℓ B B = F m I ℓ = 0 . 31 N (4 . 5 A) (0 . 45 m) = 0 . 153086 T 003 10.0 points A rectangular loop of wire hangs vertically as shown in the figure. A magnetic field is di- rected horizontally, perpendicular to the wire, and points out of the page at all points as rep- resented by the symbol . The magnetic field is very nearly uniform along the horizontal portion of the wire ab (length is 0 . 1 m) which is near the center of a large magnet produc- ing the field. The top portion of the wire loop is free of the field. The loop hangs from a balance which measures a downward force (in addition to the gravitational force) of 3 × 10 2 N when the wire carries a current 0 . 2 A. a b I I F B What is the magnitude of the magnet field B at the center of the magnet? Correct answer: 1 . 5 T. Explanation: Let : F = 3 × 10 2 , = 0 . 1 m , and I = 0 . 2 A .

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ragsdale (zdr82) – HW7 – ditmire – (58335) 2 The magnetic forces on the two vertical sec- tions of the wire loop point to the left and right, respectively. They are equal but in opposite directions, so they add up to zero. Thus the net magnetic force on the loop is that on the horizontal section ab whose length is = 0 . 1 m (and θ = 90 so sin θ = 1), and B = F I ℓ = 3 × 10 2 N (0 . 2 A) (0 . 1 m) = 1 . 5 T . 004 (part 1 of 4) 10.0 points Two wires each carry a current I in the xy plane and are subjected to an external uni- form magnetic field vector B , which is directed along the positive y axis as shown in the figure. R I wire #2 I wire #1 B L L x y What is the magnitude of the force on wire #1 due to the external vector B field? 1. bardbl vector B bardbl = (2 L + R ) I B 2. bardbl vector B bardbl = I 2 (2 L + 2 R ) B 3. bardbl vector B bardbl = parenleftbigg L + R 2 parenrightbigg I B 4. bardbl vector B bardbl = 2 I R L B 5. bardbl vector B bardbl = I ( L + R ) B 6. bardbl vector B bardbl = 2 I ( L + R ) B correct 7. bardbl vector B bardbl = 2 I L B 8. bardbl vector B bardbl = 4 I ( L + R ) B 9. bardbl vector B bardbl = (2 B L + 2 R ) I 10. bardbl vector B bardbl = 2 I R B Explanation: vector F = I integraldisplay dvectors × vector B = I vector × vector B .
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