Download PHYS 2421 Final Exam Fall 2008

PHYS 2421
Final Exam
Fall 2008
NAME___________________________________
Problem 1 (10 points). Select True or False
1. If released in an electric field, an electric dipole will move along the E-field lines. .
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2. If released in an electric field, an electric dipole will rotate increasing its potential energy. .
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3. Resistors connected in parallel suffer the same voltage drop.
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4. The capacitance of a capacitor depends only on the geometry of it .
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5. Dielectrics increase the electric field inside a capacitor.
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6. A violation of Lenz’s law would imply a violation of conservation of energy.
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7. An electron with velocity in the -x direction in a region with a B in the –y direction would feel a
magnetic force on the –z direction. .
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8. A current in the +z direction in a region with a B in the +x direction would feel a magnetic force on the
+y direction. .
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9. If the phasor corresponding to the VL is at angle of 1800, then the angle of the phasor corresponding to
the VR is at 2700
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10. All un-damped LRC circuits correspond to cases of maximum impedance .
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Problem 2 (10 points). A spherical shell with charge Q2 uniformly
distributed is concentric to a spherical charge Q1. Starting from Gauss’ law
determine expressions for the electric field at the following distances r away
from the center;
A) r < a; B) a< r < b; C) b< r < c; D) c< r
Gauss’ lawE=Qinside/4πε0r2
A) r < a: Qinside=[-Q1/4πa3/3][ 4πr3/3]= -Q1r3/a3
E=[-Q1r3/a3]/4πε0r2 = -Q1r/4πε0a3
B) a < r < b: Qinside=-Q1 E=-Q1/4πε0r2
C) b < r < c: Qinside= - Q1 +[Q2/4π( c3-b3)/3][ 4π( r3-c3)/3] = - Q1 + Q2( r3-b3)/( c3-b3)
E=[- Q1 + Q2( r3-b3)/( c3-b3)]/4πε0r2
D) c < r: Qinside= Q2-Q1 E=[ Q2-Q1]/4πε0r2
Problem 3 (10 points). In the circuit shown the switch was connected at t = 0 when
the capacitor was uncharged.
Determine A) the initial value of the current, B) The value of the charge in C after
a long time, C) The value fo the current after along time, D) the value of the
current at a time of one “time constant”
Solution:
A) i = V / R = 12 V /10 × 106 Ω = 1.2 × 10-6 A ,
B) After a long time V across C is also 12 V ⇒ Q = CV = 1× 10−6 F × 12 V = 12 × 10−6 C
C) After a long time i = 0 as the capacitor become an open circuit
D) at t = τ i=I0 e-t/ττ = (1.2x10-6 A) e-1 A
Problem 4 (10 points). At a frequency ω1, the reactance of a certain capacitor is equal to the reactance of
a certain inductor.
A) Find the ratio of the reactance of the inductor to that of the capacitor if the frequency is changed to
ω2= 2ω1?
B) If the inductor and capacitor are used with a 250 Ω resistor to construct an L-R-C circuit, at what
frequency will the impedance be minimum, and what will be the value of the impedance at such frequency?
Solution:
A) XL=XC ω1L=1/ω1C
ω1=1/(LC)1/2 XL/XC= ω2L/(1/ω2C) = ω22LC = 4 ω12LC = 4LC/LC = 4
B) Minimum impedance
XL=XC Ζ=( R + (XL - XC)2 )1/2=R=250 Ω
Problem 5 (10 points). In the circuit shown the voltage across the 2 Ω is 12 V,
A) what is the emf of the battery
B) What is the current through all other branches?
Solution: A) V2 = i2R2 i2 = V2/R2 = 12V/2Ω = 6 A i1 = 6 A
V1 = i1R1 = 6Ax1Ω= 6 V ε = 6 V+ 12 V=18 V
B) Method 1: V6 = 18 V = i6 R6 i6 = V6/R6 = 18 V/6 Ω = 3 A iε = 3 A + 6 A = 9 A
Method 2: Req = (1/(1+2)+1/6)-1= 2 Ω iε = Vε/Req = 18 V/2 Ω = 9 A
Problem 6 (10 points). A circular loop of wire with radius of 12 cm and oriented in the horizontal x-y
plane is located in a region of uniform magnetic field of 1.5 T oriented in the +Z direction.
A) What will be the average emf induced in the loop if it is removed from the field in a time interval of 2 x
10-3 s
B) Is the current induced in the loop clockwise or counterclockwise as seen from above?
Solution:
A) ε = -NdΦ /dt = - (∆Φ )/∆t = - (Φ final - Φ initial)/∆t = - (Bfinal - Binitial) πr 2/∆t
= - (0 – 1.5 T) π(0.12)2/(2x10-3 s) = 33.93 V
Flux is decreasing, B is in +Z decreasing induced field must be in +Z direction
current must be counterclockwise
EXTRA Problem 7 (10 EXTRA points). The wire carries a current I as shown. Determine
the magnitude and direction of the net magnetic field at the point P
Solution: straight segments do not contribute. Field at P is due to circular segment and
is equal to that of ¼ of circular loop:
B = (1/4)µ0I/2R = µ0I/8R