# solution6_pdf - white(rw22679 Hw06 Ross(10219 This...

• Notes
• 5
• 100% (1) 1 out of 1 people found this document helpful

This preview shows pages 1–3. Sign up to view the full content.

white (rw22679) – Hw06 – Ross – (10219) 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 long solenoid carries a current 30 A. Another coil (of larger diameter than the solenoid) is coaxial with the center of the solenoid, as in the figure below. The permeability of free space is 4 π × 10 7 N / A 2 . 8 m 1 m Outside solenoid has 270 turns Inside solenoid has 4650 turns 14 . 1 cm 4 . 1 cm Find the mutual inductance of the system. Correct answer: 0 . 00104149 H. Explanation: 2 1 Outside solenoid has N 1 turns Inside solenoid has N 2 turns A 1 A 2 Let : r 1 = 14 . 1 cm = 0 . 141 m , r 2 = 4 . 1 cm = 0 . 041 m , N 1 = 270 , N 2 = 4650 , 1 = 1 m , 2 = 8 m , n 1 = N 1 1 = 270 turns / meter , n 2 = N 1 1 = 581 . 25 turns / meter , I 2 = 30 A , and μ 0 = 4 π × 10 7 N / A 2 . M is the mutual inductance A 1 = πr 1 2 = π (0 . 141 m) 2 = 0 . 062458 m 2 A 2 = πr 2 2 = π (0 . 041 m) 2 = 0 . 00528102 m 2 B 2 = μ 0 N 2 I 2 2 The flux Φ 12 through coil 1 due to coil 2 is Φ 12 = B 2 A 2 . Thus, the mutual inductance is M = N 2 Φ 12 I 1 = N 1 Φ 21 I 2 = μ 0 N 1 N 2 A 2 I 2 2 I 2 = μ 0 N 1 N 2 A 2 2 = (4 π × 10 7 N / A 2 ) (270)(4650) 8 m × (0 . 00528102 m 2 ) = 0 . 00104149 H . 002 10.0 points A certain capacitor in a circuit has a capaci- tive reactance of 41 . 5 Ω when the frequency is 103 Hz. What capacitive reactance does the capac- itor have at a frequency of 10400 Hz? Correct answer: 0 . 41101 Ω. Explanation: Let : X C = 41 . 5 Ω , f low = 103 Hz , and f high = 10400 Hz . The capacitive reactance is X C = C 2 π f , so X C ( high ) X C ( low ) = 2 π f low C 2 π f high C = f low f high .

This preview has intentionally blurred sections. Sign up to view the full version.

white (rw22679) – Hw06 – Ross – (10219) 2 Thus X C ( high ) = X C ( low ) f low f high = (41 . 5 Ω) parenleftbigg 103 Hz 10400 Hz parenrightbigg = 0 . 41101 Ω .
This is the end of the preview. Sign up to access the rest of the document.
• Spring '10
• Rhoads
• Physics, Current, Correct Answer, Inductor, Electrical resistance

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern