# hw_ch23 - Instructor: Tomoyuki Nakayama Wednesday, June 23,...

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Instructor: Tomoyuki Nakayama Wednesday, June 23, 2010 PHY 2005 Applied Physics 2 - Summer C 2010 Solutions for Suggested Homework Problems (Chapter 23) ________________________________________________________________________________ 1. Magnetic flux through a surface is given by Φ = BA cos θ , where θ is the angle between the normal to the surface and the magnetic field. The area of the rectangular board is A = 0.2 × 0.3 = 0.06 m 2 a) Φ = BA cos0° = 9 × 10 -3 T·m 2 b) Φ = BA cos90° = 0 c) Φ = BA cos40° = 6.89 × 10 -3 T·m 2 5. The solenoid has number of turns per unit length n = N/L = 1200 turns/m. The magnetic field in the solenoid is B = k m B v = k m μ nI = 9.05 × 10 -2 T. The magnetic field is perpendicular to the cross section of the solenoid. Therefore, the magnetic flux is Φ = BA = B π ( D /2) 2 = 2.30 × 10 -5 Wb, where D is the diameter of the solenoid. 7. a) The change in flux is ΔΦ = BA (cos60°-cos0°) = -4.0 × 10 -5 Wb. Faraday’s law yields ε ind = - ΔΦ / Δ t = 8.0 × 10 -5 V. b) The change in flux is ΔΦ = BA (cos40°-cos30°) = -8.0 × 10 -6 Wb. Faraday’s law yields ε ind = - ΔΦ / Δ t = 1.6 × 10 -5 V. 9. If two coils are wound on a common ferromagnetic core, we can assume the same magnetic flux passes through both of the coils. We apply Faraday’s law to obtain

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## This note was uploaded on 12/15/2011 for the course PHY 2005 taught by Professor Lee during the Spring '08 term at University of Florida.

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hw_ch23 - Instructor: Tomoyuki Nakayama Wednesday, June 23,...

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