Physics 251 August 1, 2002

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Physics 251August 1, 2002

• Questions from yesterday?
• Review Practice Final
• Additional Problems
• Bonus Points for Formal Lab Reportssee homepage

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These notes are available on-line

• Http//webphysics.iupui.edu/251/251su02/Final_revi
  ew.ppt

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Question 1

• Q1 An ideal 10 V battery connected across a
  resistor generates 1000 J in 20 seconds. The
  value of the resistor is
• a) 1 Ohm.
• b) 2 Ohms.
• c) 50 Ohms.
• d) 100 Ohms.

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Answer

• V10 V, P1000J/20s50 W
• b) 2 Ohms

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Q2

• A wire carries current from north to south. An
  electron is located directly below the wire and
  is traveling straight up. The force on the
  electron is
• a) To the North
• b) To the South
• c) To the East
• d) To the West
• e) Zero
• f) None of the above
• 
  

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Answer

• I north to south
• v cross B points to the North, the force on a
  negative electron points opposite, or South.
• b) To the South

N
I
E
W
v
B
S
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Q3

• A series RLC circuit has XL lt XC. Which of the
  following is true?
• a) The phase is zero.
• b) The phase is negative.
• c) The power factor is
  zero.
• d) The power factor is
  positive.
• e) Two of the above.
• f) None of the above.

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answer

• XL lt XC so the circuit is more capacitor likeV
  lags I, negative phase
• The power factor is always positive
• e) Two of the above

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Q4

• The image you see when you look at yourself in
  the mirror is
• 
  a) real and erect
• 
  b) real and inverted
• 
  c) virtual and erect
• 
  d) virtual and inverted

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Answer

• c) virtual and erect

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P1

• A charge distribution consists of three charges
  located at the corners of a triangle. The charges
  are q1q2 2 microcoulombs at the points (-1,-1)
  and (1,-1) and a third charge q3 at the point
  (0,1).
• (a) Find the charge q3 that gives a zero electric
  field at the origin.
• (b) Find the total potential energy of this
  system.

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Answer

• A) By superposition, the field at the origin is
  E1E2E3. So we want E3-(E1E2). By symmetry,
  (E1E2) must be in the j direction and must have
  magnitude (E1E2)2E 1,y. If q3 is positive E2
  will automatically be in the -j direction and
  thus opposite to (E1E2). So, we need only set
  E32E1,y.

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• B) The potential energy for a pair of charges is
  Ukq1q2/r. The potential for the system must
  take into account all three pairs

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P2

• Three moles of a monatomic ideal gas begin at
  volume V 0.1 m3 and 1 atmosphere. The gas is
  taken through the following cycle.
• 1.Volume doubled at constant pressure (ab)
• 2.Pressure reduced by a factor of two at constant
  volume (bc)
• 3.Volume reduced by a factor of two at constant
  pressure (cd)
• 4.Pressure doubled at constant volume (da)
• a) Draw a P-V diagram for this cycle
• b) Calculate Q, W, and the change in internal
  energy for each of the four processes.

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Answer

• A) The pressure vs. volume diagram
• V 0.1 m3 and p1 atmosphere

a
b
p1 atm
p0.5 atm
c
d
V0.1m3
V0.2m3
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Answer b)

• Two isobaric processes, two isochoric processes
  (1 atm 1e5 Pa)

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Cont.

• To get the qs, frst calculate all the Ts using
  the ideal gas law

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Cont.

• Now, use the Ts in the equation for heat. Also
  use Cp5R/2 and CV3R/2 for monotomic ideal gas

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Cont.

• Finally, use the first law of thermodynamics
• Notice the total change in internal energy is
  zero!

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P3

• A solid insulating sphere of radius a carries a
  uniform charge density and has total charge Q.
  Concentric with this sphere is a conducting
  spherical shell with inner radius b and outer
  radius c. The conducting shell carries a total
  charge 2Q.
• a) Find the electric field in each of the
  regions r less than a, r between a and b, r
  between b and c, and r greater than c.
• b) Find the surface charge density on the inner
  and outer surfaces of the spherical shell.

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Answer

• A) By symmetry, the electric field must be
  radial, so we can use Gauss Law. Since the
  field is radial, the flux integral simplifies and
  we get

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Cont.

• To get the surface charge density on the inner
  surface, we require that a Gaussian surface drawn
  inside the conductor contain no net charge, this
  means that the charge at rb is -Q. So, at rb,
  ?-Q/4?b2. The sum of the charge at
  rc and rb must be 2Q, so there must be 3Q at
  rc. This means at rc , ?3Q/4?c2.

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P4

• Consider a series RLC circuit. The source
  supplies a voltage of V150cos(120 ?t?) volts,
  maintaining a current I I0cos(120?t). The
  values of the circuit elements are R 30 Ohms, L
  175 mH, and C 75 microfarads. Find
• a) The frequency
• b) The impedance
• c) the phase angle
• d) The average power delivered by the source.
• e) The current amplitude.

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Answer

• The frequency
• The impedance

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Cont.

• c) the phase angle

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Cont.

• d) The average power delivered by the source

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Cont.

• e) The current amplitude.

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P5

• An object is placed 30 cm to the left of lens 1,
  which has focal length f1 10 cm. Ten centimeters
  to the right of lens 1 is lens 2, which has of
  focal length f2 -10.
• a.Is lens 1 converging or diverging? How about
  Lens 2?
• b.Find the location of the final image produced
  by these two lenses together
• c.Find the total magnification of the system.
• d.Is the final image real or vitual? Erect or
  inverted?

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Answer

• A) Lens 1 is converging (fgt0). Lens 2 is
  diverging (flt0).
• B) first image due to lens 1
• with new object (virtual) distance, we have

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cont

• c.Find the total magnification of the system

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Cont.

• d.Is the final image real or vitual? Erect or
  inverted?
• The final image is real (s2gt0) and inverted
  (mTlt0)

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P6

• The figure to the right shows an apparatus
  consisting of a conducting rail that is bent into
  a circular arc of radius a 0.5 m, and a resistor
  of value R 10 Ohms that is connected to the arc
  at one end, and to a pivot point at the otherend.
  A conducting rod is connected to the resistor at
  the pivot, and slides along the rail near its
  other end. The entire apparatus is in a uniform
  magnetic field of magnitude B0 0.2 T that is
  perpendicular to the plane of the loop (into the
  page). The rod rotates around the pivot at
  constant angular velocity of 2 rad/sec. Please
  find
• a.The direction of the induced current through
  the resistor.
• b.The magnitude of the induced current.

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Answer

• A) As the rod swings, the area of the pie slice
  defined by the rod, the resistor, and the curved
  wire gets larger, and thus the flux (into the
  plane) goes too. Lenzs law says the induced
  current opposes this, so the induced current will
  flow in the direction required to produce a field
  out of the plane. Using the right hand rule,
  this direction must be from left to right in the
  resistor

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• B) The area of the pie slice is (?/2)a2 where ?
  is the angle between the rod and the resistor.
  Since the field is perpendicular to the plane,
  the flux is BAB (?/2)a2. Faradays Law gives

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Test Prep Problems