11 - kavo(ak22862 – HW11 – Janow –(12121 1 This...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: kavo (ak22862) – HW11 – Janow – (12121) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 15.0 points A plane loop of wire of area A is placed in a region where the magnetic field is perpendicu- lar to the plane. The magnitude of B varies in time according to the expression B = B e- at . That is, at t = 0 the field is B , and for t > 0, the field decreases exponentially in time. Find the induced emf, E , in the loop as a function of time. 1. E = A B e- at 2. E = a A B 3. E = a A B e- at correct 4. E = a A B e- 2 at 5. E = a B t 6. E = a B e- at Explanation: Basic Concepts: Faraday’s Law: E ≡ contintegraldisplay E · ds = − d Φ B dt Solution: Since B is perpendicular to the plane of the loop, the magnetic flux through the loop at time t > 0 is Φ B = B A = A B e- at Also, since the coefficient AB and the pa- rameter a are constants, and Faraday’s Law says E = − d Φ B dt the induced emf can be calculated the from Equation above: E = − d Φ B dt = − A B d dt e- at = a A B e- at That is, the induced emf decays exponentially in time. Note: The maximum emf occurs at t = 0 , where E = a A B . B = B e- at B vector t The plot of E versus t is similar to the B versus t curve shown in the figure above. 002 15.0 points The plane of a rectangular coil, 6 . 4 cm by 9 . 7 cm, is perpendicular to the direction of a uniform magnetic field B . If the coil has 64 turns and a total resistance of 11 Ω, at what rate must the magnitude of B change to induce a current of 0 . 09 A in the windings of the coil? Correct answer: 2 . 49174 T / s. Explanation: Given : x = 6 . 4 cm = 0 . 064 m , y = 9 . 7 cm = 0 . 097 m , N = 64 turns , r = 11 Ω , and I = 0 . 09 A . The induced emf is E = I R = N d Φ dt = N d ( B A ) dt = N A dB dt , so dB dt = I R N ( xy ) = (0 . 09 A) (11 Ω) 64(0 . 064 m) (0 . 097 m) = 2 . 49174 T / s . kavo (ak22862) – HW11 – Janow – (12121) 2 003 15.0 points A conducting bar moves as shown near a long wire carrying a constant I = 37 A current. I a v L A B If a = 9 . 3 mm, L = 45 cm, and v = 18 m / s, what is the potential difference, Δ V ≡ V A − V B ? Correct answer: 6 . 44516 mV....
View Full Document

This note was uploaded on 09/16/2009 for the course PHYS 121 taught by Professor Opyrchal during the Fall '08 term at NJIT.

Page1 / 6

11 - kavo(ak22862 – HW11 – Janow –(12121 1 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