1. Chapter E9 Symmetry and Flux E9B.1, E9B.2, E9B.5, E9B.9 and E9S.6 (only part a). Due Wednesday

2. Symmetry argument (valid for both electric and magnetic Fields. If a charge or current distribution is unchanged by some kind of a transformation (such as a rotation around an axis or sliding along an axis), then the distribution’s field is also unchanged by the same transformation.

3. Spherical distributions A spherical charge distribution is unchanged under rotation. Symmetry arguments tell us that the magnitude of the field at point S must be equal to that at point P. P S

4. cylindrical distributions A cylindrical charge distribution is unchanged under rotation. If the cylinder is infinitely long, it will also be unchanged if it slides along its axis. Symmetry arguments tell us that the magnitude of the field at point S must be equal to that at point P. P S

5. Symmetry of an infinite slab Unchanged under sliding along the axis or reflection.

6. The mirror rule for magnetic fields If we slice a current distribution with a mirror in such a way that the distribution (taking into account the direction of the current) looks exactly the same as it did before we inserted the mirror, then a nonzero magnetic field vector at any point on the mirror’s surface must be perpendicular to that surface. Consider the current in a wire. (Where do I put the mirror?)

7. A solenoid Cross section view Where do I put my mirror in this situation? What is wrong with putting my mirror here? I

8. Definitions A closed surface completely surrounds a 3 dimensional volume (with no holes). A bounded surface is surrounded by a closed curve. There can a current in the line surrounding the surface – This gives the direction the area vector points. The area vector is perpendicular to the surface, its direction given by the current around the surface. The size of the vector is given by the area.

9. Flux through an area We define a vector as being perpendicular to the surface (to a small tile placed on the surface). The magnitude of dA is proportional to the area of the tile. Electric Flux Magnetic flux

10. Flux through a closed surface This is the same as on the previous slide, except that the sum (integral) is over the entire surface, instead of an area –In a closed surface, the area vector always point to the outside. Electric Flux Magnetic flux

11. Example E 9.7 Calculate the net electric flux a through a sphere of radius r around a point charge. In this case the electric field is constant so we can remove E from the integral. The integral of dA is just the area of a sphere (4πr 2 ). The electric field for a point charge is: Note that in this case E is perpendicular to the surface at all points.

12. Magnetic units IB is the magnetic flux in relativistic units. c is the speed of light (3 x 10 8 m/s) The units of B are tesla. The units of IB are Newtons/Coulomb 1 Tesla = 1T = c N/C = 3 x 10 8 N/C = 300 MN/C (meganewtons/coulomb)

13. Example problem E9B.3 Show that E of a ring of charge can depend at most on r. r θ If E depended on θ, then when the ring was rotated, E would change in violation of the symmetry rule. E This is true also for r not in the plane of the screen.

14. Problems due Wednesday E9B.1, E9B.2, E9B.5, E9B.9 and E9S.6 (only part a).