25-1 - Kapoor (mk9499) homework25 Turner (60230) This...

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

View Full Document Right Arrow Icon
Kapoor (mk9499) – homework25 – Turner – (60230) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points An electric motor has 114 turns oF wire wrapped on a rectangular coil, oF dimensions 3 . 13 cm by 4 . 35 cm. Assume that the mo- tor uses 10 . 1 A oF current and that a uniForm 0 . 35 T magnetic feld exists within the motor. What is the maximum torque delivered by the motor? Correct answer: 0 . 548691 N m. Explanation: Let : w =3 . 13 cm , ± =4 . 35 cm , I = 10 . 1A , and B =0 . 35 T . ±orce : ² F = I ² ± × ² B Torque : ²τ = ²r × ² F Power : P = τ ω . Magnetic Force exerted on the 4 . 35 cm side is, F = N I ± B = (114 turns) (10 . 1 A) (0 . 0435 m) (0 . 35 T) = 17 . 5301 N . So, the maximum torque on the loop due to the magnetic feld is τ max = F × w = (17 . 5301 N) × (0 . 0313 m) = 0 . 548691 N m . 002 (part 2 oF 2) 10.0 points IF the motor rotates at 2740 rev / min, what is the peak power produced by the motor? Correct answer: 157 . 437 W. Explanation: P = τω = (0 . 548691 N m) (286 . 932 s - 1 ) = 157 . 437 W . 003 10.0 points A long solenoid has a coil made oF fne wire inside it and coaxial with it. I Outside solenoid has n turns per meter Inside coil has N turns r R Given a varying current I in the outer solenoid, what is the emf induced in the inner loop? 1. E = - π r 2 μ 0 nN dI dt 2. E = - π r μ 0 N dt 3. E = - 2 μ 0 n dt 4. E = - π R 2 μ 0 dt correct 5. E = - π r μ 0 n dt 6. E = - π R μ 0 N dt 7. E = - 2 μ 0 n dt 8. E = - π R μ 0 dt 9. E = - π r μ 0 dt 10. E = - π R μ 0 n dt Explanation: The magnetic feld oF a solenoid is B = μ 0 nI . The magnetic ²ux is Φ B = B · A =( μ 0 nI )( 2 ) ,
Background image of page 1

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

View Full DocumentRight Arrow Icon
Kapoor (mk9499) – homework25 – Turner – (60230) 2 so the emf is E = - d Φ B dt = - π R 2 μ 0 nN dI dt . We are interested in the emf in the inner coil, so we use the smaller area of the inner coil rather than the larger solenoid area. keywords: 004 (part 1 of 2) 10.0 points Two inFnitely long solenoids (seen in cross section in the Fgure below) thread a circuit. Given: a =0 . 9m , r 1 . 2m , r 2 . 1m , Δ B Δ t = 160 T / s , R l = 9 Ω , R m = 4 Ω , and R r =2 . 6 Ω , as in the Fgure below. The magnitude of B inside each solenoid is the same and is 300 T at time t = 0. B in B out I l I m I r r 2 R m R l R r r 1 a a a What is the magnitude of the current, I m , in middle resistor, R m ? Correct answer: 1 . 39706 A. Explanation: Basic Concept: ±araday’s Law: E = - d Φ B dt Ohm’s Law: I = V R Junction Rule: n ± i =1 I i Solution: Note: The side-length, a , of the circuit loop is not necessary for this problem. Neither is the magnitude of B at time t = 0. ±rom ±araday’s law, the induced emf in the left loop (Loop 1) is |E 1 | = d Φ B dt = A 1 dB dt = π r 2 1 dt = 20 . 1062 V and the direction of this emf is counter- clockwise. Similarly, the induced emf in the right loop (Loop 2) is |E 2 | = d Φ B dt = A 2 dt = 2 2 dt =5 . 02655 V and the direction of this emf is clockwise. (Numerical values are inserted into the equations to verify answers; our calculation has only 7 place accuracy.) Moving clockwise around the loops, circuit equations for these two loops are Loop 1: - R l I l + R m I m -E 1 = 0 (1) - (9 Ω)( - 1 . 6131 A) +(4 Ω)(1 . 39706 A) - 20 . 1062 V = 0 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

25-1 - Kapoor (mk9499) homework25 Turner (60230) This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online