1 μ 2 π ir r 2 2 μ 2 π ir 2 r 3 3 μ 4 π ir 2 r

This preview shows page 3 - 6 out of 6 pages.

1. μ 0 2 π IR r 2 2. μ 0 2 π IR 2 r 3 3. μ 0 4 π Ir 2 R 3 4. μ 0 4 π Ir 2 R 5. μ 0 2 π Ir 2 R 3 6. μ 0 4 π IR 2 r 3 7. μ 0 4 π Ir R 2 8. μ 0 4 π IR r 2 9. μ 0 2 π Ir 2 R 10. μ 0 2 π Ir R 2 correct Explanation: The current density (current/area) is the same throughout the wire. Sketch an Ampe- rian loop within the wire with radius r . Then, the current through this loop i divided by area of this loop is equal to the total current per unit area. i π r 2 = I π R 2 i = I r 2 R 2 The magnetic field is tangential to the Am- perian loop. Apply Ampere’s law by inte-
Image of page 3

Subscribe to view the full document.

forsythe (acf2464) – HW8-c – li – (55020) 4 grating counterclockwise around the loop if looking at the loop from the right end. Then the current flows out of the surface and is positive. contintegraldisplay vector B · d vector = μ 0 I inside path B (2 π r ) = μ 0 I r 2 R 2 B = μ 0 2 π I r R 2 . 006 10.0points An irregularly shaped metal object carries a net charge + Q . Due to its irregular shape, the surface charge density on the surface of the metal σ varies as a function of position ( σ = σ ( x, y, z ) is not a constant value). Consider a point P = ( x 0 , y 0 , z 0 ) on the surface of the object. What will be the magnitude of the electric field at point a Q very close to P but just outside the surface of the object? Hint: Imagine Q to be so close to the surface that for an observer at Q , the surface of the object looks like a large flat plane with nearly uniform charge density σ ( x 0 , y 0 , z 0 ) = σ 0 . 1. σ 0 ǫ 0 correct 2. 0 3. σ 0 2 ǫ 0 4. Depends upon the shape of the object 5. 4 σ 0 ǫ 0 6. 2 σ 0 ǫ 0 7. σ 0 4 ǫ 0 Explanation: The easiest way to do this problem is to use Gauss’s law along with the hint given. Con- sider a Gaussian surface in the shape of a very short cylinder (radius height) whose axis is perpendicular to the surface of the object and passes through P . The upper circular surface contains the point Q , while the lower circular surface is inside the metal object (while the curved surface is partly inside and partly out- side). If A is the area of the circular surface, then the net charge enclosed by this Gaussian surface will be σ 0 A , since all the charge will be present on the surface of the metal object. For calculating the electric flux, the curved surface will not contribute because the elec- tric field will be perpendicular to the surface of the metal object (and hence parallel to the curved surface). The inner surface will not contribute because the net electric field inside the metal must be zero. If E is the electric field at point Q , then the net flux will be EA . From Gauss’s law, we get EA = σ 0 A ǫ 0 E = σ 0 ǫ 0 007(part1of3)10.0points Consider two parallel metal plates of area A separated by a distance d as shown in Fig. A. The left plate carries total charge 2 Q , while the right plate carries total charge Q . As- sume that the plates are very large relative to the separation distance so that edge ef- fects may be neglected. Also assume that the thickness of the plates is finite but negligible.
Image of page 4
forsythe (acf2464) – HW8-c – li – (55020) 5 Use the superposition principle to deter- mine the E -field in regions I, II, and III. To
Image of page 5

Subscribe to view the full document.

Image of page 6
  • Spring '08
  • Turner
  • Electrostatics, Magnetic Field, Electric charge, A. Gauss

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