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Old Midterm 3

Old Midterm 3 - benavides(jjb2356 oldmidterm 03...

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benavides (jjb2356) – oldmidterm 03 – Turner – (59130) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A copper strip (8 . 47 × 10 22 electrons per cu- bic centimeter) 13 . 7 cm wide and 0 . 03 cm thick is used to measure the magnitudes of unknown magnetic fields that are perpendic- ular to the strip. The charge on the electron is 1 . 6 × 10 19 C. Find the magnitude of B when the current is 22 A and the Hall voltage is 2 . 8 μ V. Correct answer: 0 . 51744 T. Explanation: Let : n = 8 . 47 × 10 22 cm 3 , = 8 . 47 × 10 28 m 3 , q = 1 . 6 × 10 19 C , t = 0 . 03 cm = 0 . 0003 m , w = 13 . 7 cm = 0 . 137 m , I = 22 A , and V H = 2 . 8 μ V = 2 . 8 × 10 6 V . The current in the metal strip is I = n q v d A = n q v d ( w t ) v d w = I n q t The Hall voltage is V H = v d w B B = V H v d w B = n q t V H I = (8 . 47 × 10 28 m 3 ) (1 . 6 × 10 19 C) 22 A × (0 . 0003 m) (2 . 8 × 10 6 V) = 0 . 51744 T . 002 10.0 points A small rectangular coil composed of 23 turns of wire has an area of 47 cm 2 and carries a current of 0 . 6 A. When the plane of the coil makes an angle of 30 with a uniform magnetic field, the torque on the coil is 0 . 14 N m. What is the magnitude of the magnetic field? Correct answer: 2 . 49242 T. Explanation: Let : N = 23 turns , I = 0 . 6 A , θ = 30 , A = 47 cm 2 = 0 . 0047 m 2 , and τ = 0 . 14 N m . The magnetic force on the current is vector F = I vector × vector B and the torque is vector τ = vector r × vector F , so the torque on the loop due to the magnetic field is τ = 2 F r cos θ = ( N I ℓ B ) w cos θ = N I B ( ℓ w ) cos θ = N I B A cos θ , where A is the area of the loop and θ is the angle between the plane of the loop and the magnetic field. The magnetic field from above is B = τ N I A cos θ = 0 . 14 N m (23 turns) (0 . 6 A) (0 . 0047 m 2 ) cos(30 ) = 2 . 49242 T . 003 10.0 points Given: Assume the bar and rails have neg- ligible resistance and friction.

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benavides (jjb2356) – oldmidterm 03 – Turner – (59130) 2 In the arrangement shown in the figure, the resistor is 6 Ω and a 9 T magnetic field is directed out of the paper. The separation between the rails is 5 m . Neglect the mass of the bar. An applied force moves the bar to the left at a constant speed of 8 m / s . m 1 g 8 m / s 6 Ω 9 T 9 T I 5 m At what rate is energy dissipated in the resistor? Correct answer: 21600 W. Explanation: Basic Concept: Motional E E = B ℓ v . Ohm’s Law I = V R . Solution: The motional E induced in the circuit is E = B ℓ v = (9 T) (5 m) (8 m / s) = 360 V . From Ohm’s law, the current flowing through the resistor is I = E R = B ℓ v R = (9 T) (5 m) (8 m / s) R = 60 A . The power dissipated in the resistor is P = I 2 R = B 2 2 v 2 R 2 R = B 2 2 v 2 R = (9 T) 2 (5 m) 2 (8 m / s) 2 (6 Ω) = 21600 W . Note: Third of four versions. 004 10.0 points A coil is wrapped with 809 turns of wire on the perimeter of a circular frame (of radius 73 cm). Each turn has the same area, equal to that of the frame. A uniform magnetic field is directed perpendicular to the plane of the coil. This field changes at a constant rate from 27 mT to 48 mT in 61 ms.
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