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Unformatted text preview: PHY2049 Fall11 Final Final Exam Solutions 1. Three charges form an equilateral triangle of side length d = 2 cm. The top charge is q3 = 3 μC, while the bottom two are q1 = q2 =  6 μC. What is the magnitude of the net force acting on q3? (1) 700 N (2) 350 N (3) 810 N (4) 405 N (5) 0 N By symmetry the net force will be aligned vertically downward (attractive force). So we need to project the force from each of the two lower forces onto the vertical axis: qq
qq
Fnet = k 1 2 3 cos 30 + k 2 2 3 cos 30
d
d
−6
3 × 10 ) ( 6 × 10 −6 ) 3
9(
= 2 ( 9 × 10 )
= 700
2
( 0.02 )2 2. A glass rod forms a semi circle of radius r = 4 cm with a charge of  q distributed uniformly along the right quadrant and +q distributed along the left quadrant, where q = 5 pC. What is the magnitude and the direction (as the polar angle relative to the direction of the x  axis) of the electric field at the center P of the semi circle. (1) 29 N/C, θ = 0° (2) 45 N/C, θ = 270° (3) 0 N/C (4) 29 N/C, θ = 90° (5) 45 N/C, θ = 180° PHY2049 Fall11 Final The diagram shows the direction of the electric field contribution from each i
quadrant. By symmetry, the net field points down in the ˆ direction, which is θ = 0 The magnitude (not really needed to choose the correct answer!) is given by: q
λ=
π
R
2
dq = λ ds = λ Rdθ
π
π
dq
λ Rdθ
+
Ey q = ∫ k 2 sin θ = ∫ k
sin θ
π /2 R
π /2
R2
λ
λ
= − k cosθ π /2 = k
π
R
R
λ 4k q
t
⇒ Eyot = 2 k =
= 29
R π R2 3. A parallel plate capacitor with capacitance 0.5μF is connected to a 6V battery. If the plates are squeezed to half of their original separation, what will be the charge stored on the capacitor? (1) 6μC (2) 3μC (3) 1.5μC (4) 0.5μC (5) 12μC A
The capacitance of a parallel plate capacitor is C = ε 0 , so the capacitance d
increases two times when the plates are squeezed together. So the new capacitance is 1 μF. The charge is thus q = CV = 6 µC PHY2049 Fall11 Final y + + + + + + + E e x − − − − − − 4. A beam of electrons (“cathode rays”) is sent between two parallel electric plates with an electric field between them of E =  2x10^4 N/C j hat N/C . If the electron beam travels perpendicular to the electric field with a velocity of v = 4.2x 10^7 m/s in the +i  hat direction, what magnetic field is necessary (direction and magnitude) so that the electrons continue traveling in a straight line
without deflection by the electric field? Balance forces: Replace k with –k below: F = q(E + v × B ) = 0 ⇒E+ v×B = 0
Eˆ
⇒ B = k so that ˆ × k = − ˆ
jˆ
i
v E = 2 × 104 N/C
v = 4.2 × 107 m/s
⇒B= 2 × 104 N/C
= 0.48 mT
4.2 × 107 PHY2049 Fall11 Final 5. Part of a long insulated wire carrying current i = 4A is bent into a circular section of radius R = 1.5 cm as shown in the figure. What is the magnitude of the magnetic field (in T) at the center of curvature C if the circular section lies in the plane of the page as shown? (1) 2.2 x 10 4 (2) 5.3 x 10 5 (3) 1.7 x 10 4 (4) 1.1 x 10 4 (5) 5.0 x 10 6 The field from the long straight wire is given by µi
Bw = 0 out of the page at the center of the circle. 2π R The field from the circular loop itself is given by the Biot Savart law: µi
µi
Bloop = 0 φ = 0 also out of the page 4π R
2R Both contributions add, given a field B=2.2 x 10 4 6. A series RLC circuit is driven by a sinusoidally varying EMF source. The current lags the EMF by 30°. What can be concluded about the driving frequency ω? 1
1
1
(1) ω > (2) ω < (3) ω = (4) ω = 0 (5) ω = ∞ LC
LC
LC ε ( t ) = ε m sin (ω t ) i ( t ) = im sin (ω t − φ ) So the current lags the EMF when Φ>0 X − XC
tan φ = L
> 0 R X L > XC
1
ωL >
ωC 1
ω2 >
LC
1
⇒ω >
LC PHY2049 Fall11 Final 7. A constant current of i = 3A is used to charge a parallel plate capacitor with circular plates of radius R = 1cm. What is the magnitude of the magnetic field at a radius of r = 0.3 cm, which is less than R, in the region between the plates? (1) 1.8 x 10 5 T (2) 6 x 10 5 T (3) 1.2 x 10 4 T (4) 3.8 x 10 6 T (5) 0 T The total displacement current between the plates is id=i=3A. The effective i
displacement current density is jd = d 2 πR
So using Maxwell’s Law of Induction: B ⋅ ds = µ0id′
∫ 2π rB = µ0 jdπ r 2 = µ0
B= id
π r 2 π R2 µ0 id
r = 1.8 × 10 −5 T
2π R 2 8. Light traveling horizontally enters a right prism as shown in the figure. The index of refraction of the prism is n=1.6, and it is surrounded by air. What is the measure of the angle θ2 that the light deflected from horizontal when it exits the prism? 45° θ1 θ2 (1) 31° (2) 26° (3) 19° (4) 45° (5) 0° There are 2 refractions from the 2 surfaces. And since the second surface is not parallel to the first, the light does not return to the incident direction. PHY2049 Fall11 Final sin 45 = 1.6sin θ1 1.6sin ( 45 − θ1 ) = sin θ 2 ⇒ θ 2 = 31o 9. An object
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This note was uploaded on 01/03/2012 for the course PHY 2049 taught by Professor Any during the Fall '08 term at University of Florida.
 Fall '08
 Any
 Physics, Charge

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