This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: nguyen (jmn727) oldmidterm 03 Turner (59070) 1 This printout should have 19 questions. Multiplechoice 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 . 4 cm wide and 0 . 15 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 23 A and the Hall voltage is 2 V. Correct answer: 1 . 76765 T. Explanation: Let : n = 8 . 47 10 22 cm 3 , = 8 . 47 10 28 m 3 , q = 1 . 6 10 19 C , t = 0 . 15 cm = 0 . 0015 m , w = 13 . 4 cm = 0 . 134 m , I = 23 A , and V H = 2 V = 2 10 6 V . The current in the metal strip is I = nq v d A = nq v d ( w t ) v d w = I nq t The Hall voltage is V H = v d w B B = V H v d w B = nq tV H I = (8 . 47 10 28 m 3 ) (1 . 6 10 19 C) 23 A (0 . 0015 m) (2 10 6 V) = 1 . 76765 T . 002 10.0 points A small rectangular coil composed of 75 turns of wire has an area of 25 cm 2 and carries a current of 1 . 6 A. When the plane of the coil makes an angle of 38 with a uniform magnetic field, the torque on the coil is 0 . 06 N m. What is the magnitude of the magnetic field? Correct answer: 0 . 253804 T. Explanation: Let : N = 75 turns , I = 1 . 6 A , = 38 , A = 25 cm 2 = 0 . 0025 m 2 , and = 0 . 06 N m . The magnetic force on the current is vector F = I vector vector B and the torque is vector = vector r vector F , so the torque on the loop due to the magnetic field is = 2 F r cos = ( N I B ) w cos = N I B ( w ) cos = N I B A cos , where A is the area of the loop and is the angle between the plane of the loop and the magnetic field. The magnetic field from above is B = N I A cos = . 06 N m (75 turns) (1 . 6 A) (0 . 0025 m 2 ) cos(38 ) = . 253804 T . 003 10.0 points Given: Assume the bar and rails have neg ligible resistance and friction. nguyen (jmn727) oldmidterm 03 Turner (59070) 2 In the arrangement shown in the figure, the resistor is 9 and a 4 T magnetic field is directed out of the paper. The separation between the rails is 2 m . Neglect the mass of the bar. An applied force moves the bar to the right at a constant speed of 7 m / s . m 1g 7 m / s 9 4 T 4 T I 2m At what rate is energy dissipated in the resistor? Correct answer: 348 . 444 W. Explanation: Basic Concept: Motional E E = B v . Ohms Law I = V R . Solution: The motional E induced in the circuit is E = B v = (4 T) (2 m) (7 m / s) = 56 V . From Ohms law, the current flowing through the resistor is I = E R = B v R = (4 T) (2 m) (7 m / s) R = 6 . 22222 A ....
View
Full
Document
 Spring '08
 Turner

Click to edit the document details