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solution11_pdf

# solution11_pdf - maini(nm7637 hw11 Shneidman(12108 This...

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maini (nm7637) – hw11 – Shneidman – (12108) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A plane loop oF wire oF area A is placed in a region where the magnetic feld 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 feld is B 0 , and For t > 0, the feld decreases exponentially in time. ±ind the induced emF, E , in the loop as a Function oF time. 1. E = A B 0 e - at 2. E = a B 0 e - at 3. E = a A B 0 e - at correct 4. E = a A B 0 5. E = a A B 0 e - 2 at 6. E = a B 0 t Explanation: Basic Concepts: ±araday’s Law: E ≡ c E · ds = d Φ B dt Solution: Since B is perpendicular to the plane oF the loop, the magnetic ²ux through the loop at time t > 0 is Φ B = B A = A B 0 e - at Also, since the coe³cient AB 0 and the pa- rameter a are constants, and ±araday’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 - = a A B 0 e - 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 v t The plot oF E versus t is similar to the B versus t curve shown in the fgure above. 002 10.0 points The plane oF a rectangular coil, 4 . 8 cm by 6 . 8 cm, is perpendicular to the direction oF a uniForm magnetic feld B . IF the coil has 45 turns and a total resistance oF 11 . 3 Ω, at what rate must the magnitude oF B change to induce a current oF 0 . 05 A in the windings oF the coil? Correct answer: 3 . 84668 T / s. Explanation: Given : x = 4 . 8 cm = 0 . 048 m , y = 6 . 8 cm = 0 . 068 m , N = 45 turns , r = 11 . 3 Ω , and I = 0 . 05 A . The induced emF is E = I R = N d Φ dt = N d ( B A ) dt = N A d B dt , so d B dt = I R N ( x y ) = (0 . 05 A) (11 . 3 Ω) 45(0 . 048 m) (0 . 068 m) = 3 . 84668 T / s .

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maini (nm7637) – hw11 – Shneidman – (12108) 2 003 10.0 points A conducting bar moves as shown near a long wire carrying a constant I = 40 A current. I a v L A B If a = 8 mm, L = 120 cm, and v = 15 m / s, what is the potential diFerence, Δ V V A V B ? Correct answer: 18 mV. Explanation: Given; E = d Φ B dt = d dt ( B ℓ x ) = B ℓ dx dt = B ℓ v . ±rom Ampere’s law, the strength of the magnetic ²eld created by the long current- carrying wire at a distance a from the wire is B = μ 0 I 2 π a .
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solution11_pdf - maini(nm7637 hw11 Shneidman(12108 This...

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