{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw 24 - nguyen(jmn727 homework 24 Turner(59070 This...

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

nguyen (jmn727) – homework 24 – Turner – (59070) 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 10.0 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 0 e - at . That is, at t = 0 the field is B 0 , 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 A B 0 e - 2 at 2. E = a A B 0 3. E = A B 0 e - at 4. E = a B 0 e - at 5. E = a A B 0 e - at correct 6. E = a B 0 t 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 0 e - at Also, since the coefficient AB 0 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 0 d dt e - at = a A B 0 e - at That is, the induced emf decays exponentially in time. Note: The maximum emf occurs at t = 0 , where E = a A B 0 . B = B 0 e - at B 0 0 0 vector t The plot of E versus t is similar to the B versus t curve shown in the figure above. 002 10.0 points The magnetic flux threading a metal ring varies with time t according to Φ B = 3 a t 3 - b t 2 , with a = 4 . 7 s - 3 · m 2 · T, and b = 3 . 1 s - 2 · m 2 · T. The resistance of the ring is 2 . 5 Ω. Determine the maximum current induced in the ring during the interval from t 1 = - 4 s to t 2 = 3 s. Correct answer: 0 . 0908747 A. Explanation: From Faraday’s law, the induced emf should be E = - d Φ B dt = - (9 a t 2 - 2 b t ) , so the maximum E occurs when d E dt = - 18 a t + 2 b = 0 t = b 9 a and the maximum emf is E max = - 9 a parenleftbigg b 9 a parenrightbigg 2 + 2 b parenleftbigg b 9 a parenrightbigg = - b 2 9 a + 2 b 2 9 a = b 2 9 a .

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

View Full Document
nguyen (jmn727) – homework 24 – Turner – (59070) 2 Thus the maximum current is I max = E max R = b 2 9 a R = (3 . 1 s - 2 · m 2 · T) 2 9 (4 . 7 s - 3 · m 2 · T) (2 . 5 Ω) = 0 . 0908747 A .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 6

hw 24 - nguyen(jmn727 homework 24 Turner(59070 This...

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

View Full Document
Ask a homework question - tutors are online