11 - kavo(ak22862 HW11 Janow(12121 This print-out should...

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

View Full Document Right Arrow Icon
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 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 B 0 e - at 2. E = a A B 0 3. E = a A B 0 e - at correct 4. E = a A B 0 e - 2 at 5. E = a B 0 t 6. E = a B 0 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 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 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 d B dt , so d B dt = I R N ( x y ) = (0 . 09 A) (11 Ω) 64 (0 . 064 m) (0 . 097 m) = 2 . 49174 T / s .
Image of page 1

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

View Full Document Right Arrow Icon
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. Explanation: Given; E = d Φ B dt = d dt ( B ℓ x ) = B ℓ dx dt = B ℓ v . From Ampere’s law, the strength of the magnetic field created by the long current- carrying wire at a distance a from the wire is B = μ 0 I 2 π a .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern