1.1 × 1020 electrons flow through a cross section of a 2.4-mm-diameter iron wire in 6.0 s. Part A What is the electron drift speed? Express your answer in micrometers per second. ▸ View Available Hint(s) V ΕΠΙ ΑΣΦ V-7.90 Submit Previous Answers × Incorrect; Try Again ? μm/s

sq 4. Maximum surface resistance. Consider a square sheet of side L, thickness d, and electrical resistivity p. The resistance measured between opposite edges of the sheet is called the surface resistance: R = pL/ Ld = p/d, which is independent of the area of the sheet. (R is called the resistance per square and is expected in ohms per square, because p/d has the dimensions of ohms.) If we express p by p=m/ne²t, then Rq=m/ndet. Suppose now that the minimum value of the collision time is determined by scattering from surfaces of the sheet, so that Td/v, where v is the Fermi velocity. Thus the maximum surface resistivity is Rmvp/nd²e². Show for a monatomic metal sheet one atom thickness that sq Rħ/e² = 4.1k, where lkn is 10³ ohms. sq

1.3x1016 electrons flow through a cross section of silver wire in 310 us with a drift speed of 7.2x10-4 m/s. ▼ Part A What is the diameter of the wire? Express your answer to two significant figures and include the appropriate units. D = Submit μÅ Value Request Answer Ć Units ?

This homework problem is giving me some issues, help would be greatly appreciated! Thanks!