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Department
Physics
Course
PHY 303L
Professor
Storr
Semester
Summer

Description
rashid (sar3553) – HW10 InductanceandRLCCircuits – storr – (37537) 1 This print-out should have 16 questions. Compute the magnetic flux through each Multiple-choice questions may continue on turn. the next column or page – find all choices before answering. Correct answer: 2.34533 × 10 Wb. Explanation: 001 (part 1 of 3) 10.0 points A 37.6 mA current is carried by a uniformly wound air-core solenoid with 482 turns as Let : B = 0.000210873 T, and shown in the figure below. d = 11.9 mm = 0.0119 m. 10.8 cm The magnetic flux through each turn is π 2 Φ = B A = B 4 d 11.9 mm π 2 . mA = (0.000210873 T) (0.0119 m) 37 4 = 2.34533 × 10−8Wb . Compute the magnetic field inside the solenoid. The permeability of free space is 003 (part 3 of 3) 10.0 points −7 −6 2 Compute the inductance of the solenoid. 4π × 10 T · m/A = 1.25664 × 10 N/A . Correct answer: 0.000210873 T. Correct answer: 0.300652 mH. Explanation: Explanation: The inductance of the solenoid is Let : N = 482, L = N Φ I ℓ = 10.8 cm, (482)(2.34533 × 108 Wb) 1000 mH I = 37.6 mA, and = −6 2 0.0376 A 1 H ▯0= 1.25664 × 10 N/A . = 0.300652 mH . ℓ 004 (part 1 of 2) 10.0 points A solenoid has 130 turns of wire uniformly wrapped around an air-filled core, which has d I a diameter of 10 mm and a length of 5.4 cm. The permeability of free space is 1.25664 × 106 N/A . Calculate the self-inductance of the The magnetic field inside the solenoid is solenoid. N B = ▯ 0I = ▯ 0 I −5 ℓ Correct answer: 3.08882 × 10 H. Explanation: = (1.25664 × 106 N/A ) 482 0.108 m × (0.0376 A) Let : N = 130, = 0.000210873 T . D = 10 mm = 0.01 m, ℓ = 5.4 cm = 0.054 m, and −6 2 002 (part 2 of 3) 10.0 points ▯ 0 1.25664 × 10 N/A . rashid (sar3553) – HW10 InductanceandRLCCircuits – storr – (37537) 2 The self-inductance of a solenoid is given by 006 (part 1 of 2) 10.0 points N Φ A long insulated wire with a resistance of L 1 I , 15 Ω/m is to be used to construct a resistor. where Φ is the total flux inside the solenoid First, the wire is bent in half and then the and I is the current in the wire wrapped doubled wire is wound in a cylinderical form, around the solenoid. as show in figure. The diameter of the cylin- By Ampere’s Law, the magnetic field in the drical form is 2 cm, its length is 30 cm, and solenoid is the total length of wire is 15 m. ▯ 0 I B = ℓ , where ▯ = ▯ 0s the magnetic permeability of the air core which is the same as free space. The magnetic flux in the solenoid is 2 Φ = B A = B π D . 4 Find the inductance of this wire-wound reisitor. Using the above expressions for Φ and B, we obtain for the inductance of the solenoid Correct answer: 0 H. N B A L1= Explanation: I ▯ 0 π D 2 = Let : ℓ = 30 cm = 0.3 m, 4ℓ −6 2 2 = (1.25664 × 10 N/A )(130) d = 2 cm = 0.02 m, 2 × π (0.01 m) r = 15 Ω/m, and 4(0.054 m) L = 15 m. = 3.08882 × 10 −5H . The total flux in the coil is zero, so L = 0 H . 005 (part 2 of 2) 10.0 points The core is replaced with a soft iron rod that has the same dimensions, but a magnetic per- 007 (part 2 of 2) 10.0 points meability of 800▯0. Find the resistance of this wire-wound resis- What is the new inductance? tor. Correct answer: 0.0247106 H. Correct answer: 225 Ω. Explanation: Explanation: Let : ▯ = 800▯ 0 Let : r = 15 Ω/m and L = 15 m. ▯N π D 2 L2= The total resistance of the wire is 4ℓ = ▯ L ▯ 0 1 R = r L −5 = (15 Ω/m)(15 m) = 800(3.08882 × 10 H) = 225 Ω . = 0.0247106 H . rashid (sar3553) – HW10 InductanceandRLCCircuits – storr – (37537) 3 magnetic flux, which in turn is proportional keywords: to the rate of change of the current. This is expressed as 008 10.0 points dI A 2.33 H inductor carries a steady current E = L of 0.556 A. When the switch in the cir- dt cuit is thrown open, the current disappears in = L d (bt − at) 10.6 ms. dt
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