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: benavides (jjb2356) – oldmidterm 03 – Turner – (59130) 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 . 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 = 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) 22 A × (0 . 0003 m) (2 . 8 × 10 − 6 V) = . 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 ℓ × 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 = . 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. 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 ≪ 1g 8 m / s 6Ω 9 T 9 T I 5m 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 ....
View
Full
Document
This note was uploaded on 03/26/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

Click to edit the document details