No Title
PHYSICS III
Solution to Exam 2
                                                                                                           
Thursday 4/14/16 Name
O205: Periods 1, 2
JFW CM:
Period:














                    
                                         (1)
(25)
(2)
(25)
(3)
(25)
(4)
(25)
TOTAL
(100)
  1. A triangular coil of N equal turns lies in the x-y plane with one corner located at the origin. The total resistance of the coil is R and the coil is bathed in a spatially uniform, time dependent, magnetic field at an angle f with the z-axis.

    N=14 turns
    a=30.0 cm
    b=40.0 cm
    R=2.40 W
    B(t)=0.600 T-(0.250 T/s)t
    f = 24.0o

      (a) [10 pts] Calculate the total magnetic flux passing the coil (equation only).

      FB
      =
      NBAcosf
      =
      (14 turns)[0.600 T-(0.250 T/s)t](0.500)(0.300 m)(0.400 m) cos24.0o
      =
      0.460 Wb-(0.192 Wb/s)t

      (b) [10 pts] Calculate the current induced in each turn of the coil.

      Eind
      =
      - dFB
      dt
      =
      +0.192 Wb/s
      =
      +0.192 V
      iind
      =
      Eind
      R
      =
      0.192 V
      2.40 W
      =
      7.79×10-2 V
      �
      77.9 mV

      (c) [5 pts] In the diagram, draw in the direction of Bind. Also, looking in along the z-axis towards the origin, draw in the direction of the current flow.
      ANS: CCW
  2. A constant, external mechanical force F=Fy [^(j)] is applied to a slide wire segment [`ab] causing it to move with constant speed, v. The slide wire contacts a U-shaped wire having resistance R to form a complete electrical circuit. The system is bathed in a uniform magnetic field, B=+B [^(k)]. The constants are
    R=0.100 W
    Lab=15.0 cm
    B=0.750 T
    v=2.40 m/s

      (a) [5 pts] Which point, a or b, is at the higher electrical potential?
      ANS: Point b is at the higher potential. Use the right-hand-rule.
      (b) [5 pts] Calculate the induced emf in the slide wire.

      Eind
      =
      BvLab
      =
      (0.750 T)(2.40 m/s)(0.150 m)
      =
      0.270 V

      (c) [5 pts] Calculate the magnitude of the induced electric field in the slide wire.

      E
      =
      E
      Lab
      =
      0.270 V
      0.150 m
      =
      1.80 V/m

      (d) [5 pts] Calculate the magnitude of the mechanical force applied to the slide wire.

      Fmech
      =
      ILabB
      =
      ELabB
      R
      =
      (0.270 V)(0.150 m)(0.750 T)
      0.100 W
      =
      0.304 N

      (e) [5 pts] Calculate the power dissipated in the slide wire.
      P
      =
      Fmechv
      =
      (0.304 N)(2.40 m/s)
      =
      0.729 W
  3. The diagram shows a single turn circular wire symmetrically placed inside a single turn square coil (both lie in the same plane). The mutual inductance of the combination is M. A time-dependent current cource powers the square wire. A switch is also located in the square wire.
    M=8.60×10-3 H
    i(t)=(4.80 A)e-0.200t

      (a) [8 pts] At t=0, switch S is closed. At what rate is the current changing in the square wire at t=0.120 s?

      di1
      dt
      =
      (4.80 V)(-0.200/s)e-(0.200/s)(0.120 s)
      =
      -0.937 A/s

      (b) [7 pts] Calculate the induced emf in the circular wire at t=0.120 s.

      E2
      =
      -M21 di1
      dt
      =
      -(8.60×10-3 H)(-0.937 A/s)
      =
      8.06×10-3 V
      �
      8.06 mV

      (c) [4 pts] In the figure, draw in the direction of the induced current in the circular wire.
      ANS: CCW
      (d) [6 pts] If the current source and switch are now removed from the square wire and inserted into the circular wire and the process in part (a) is repeated, what are the induced voltage and the corresponding current direction in the square wire at t=0.120 s?
      ANS: E2=8.06 mV  CCW The direction of the induced electric field is CCW everwhere.
  4. The figure shows a simple circuit containing a battery, switch, resistor and, a REAL inductor.
    Eb=95.0 V
    R=30.0 W
    RL=50.0 W
    L=8.00 mH

      (a) [5 pts] Calculate the value of the inductive time constant, tL.

      tL
      =
      L
      Rtot
      =
      8.00×10-3 H
      30.0 W+ 50.0 W
      =
      1.00×10-4 s

      (b) [5 pts] When the switch is just closed at t=0, what is the magnitude of the induced emf in the inductor?
      ANS: 95.0 V
      (c) [5 pts] At t=5.00×10-5 s, calculate the net current in the circuit.

      i(t)
      =
      If(1-e-t/tL)
      =
      �
      �       		95.0 V
      80.0 W
      		 �
      �       		(1-e-(5.00× 10-5 s)/(1.00×10-4 s))
      =
      0.467 A

      (d) [5 pts] At t=5.00×10-5 s, calculate the induced emf in the inductor.

      E
      =
      iR+EL
      EL
      =
      95.0 V-(0.467 A)(80.0 W)
      =
      57.6 V

      (e) [5 pts] At t=5.00×10-5 s, at what rate is the net current changing?
      EL
      =
      -L di
      dt
      di
      dt
      =
      - E
      L
      =
      - 57.6 V
      8.00×10-3 H
      =
      -7.20×103 A/s
FORMULAS

E=-N dFB/dt
FB= �
�

S 
 B·
^
n
 
   dA
V=IR
UL=Li2/2
i=if(1-e-t/tL)
N2F2=M21i1
E2=-M21 di1/dt
tL=L/R
E=BLv
E=EL
P=F·v
P=E2/R
F=IL×B
E=-Ldi/dt


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On 27 Apr 2016, 10:31.