Download PHY2054 Exam II, Fall, 2011 Solutions 1.) A 5 kΩ resistor in series

PHY2054 Exam II, Fall, 2011
Solutions
1.) A 5 kΩ
Ω resistor in series with an uncharged capacitor ‘C’ is connected to a 9 V battery.
3 seconds after the connection, the voltage across the capacitor is 3 V. What is C in units of
mF? (1 mF=10-3 F).
a. 1.48
b. 0.55
c. 0.50
d. 0.72
e. 0.97
soln.: use the charging equation, q=Q(1-e-t/RC), where Q is Qmax and Q=CV, where have –
dividing by C on both sides to get the voltage across the capacitor as a function of time:
V=Vmax(1-e-t/RC), so we have 3 V=9 V(1-e-3/5000C), so 1/3 = (1-e-3/5000C), or 2/3 = e-3/5000C
take the natural log of both sides, so -0.4055=-3/5000 C or C=1.48 mF
2. If the voltage between A and B in the figure is 12 V, how much current, in A, flows thru
the 2 Ω resistor?
a. 0.81
b. 0.61
c. 0.71
d. 0.91
e. 1.01
soln.: in units of Ω, Req = 1 + (1/2 + 1/3)-1 + 4 + (1/5 + 1/6)-1 = 1 + 1.2 +4 + 2.73 Ω = 8.93 Ω
voltage drop across the 2 and 3 Ω resistors in parallel is (1.2/8.93)*12 V = 1.61 V, so V=IR,
or I thru the 2 Ω resistor is 1.61 V/2 Ω = 0.81 A
3. What is the voltage difference between A and
B, in V? (The ground, with zero potential, is at
the lower left corner of the circuit.)
a. 1.57
b. 0.57
0.57 e. 10.43
c. 1.07
d. -
Soln.: left loop, start at lower left corner to the
left of the 7 Ω resistor and go clockwise (CW).
Pick i1 going CW as the current out of the 9 V
battery flowing to point A, pick i2 going right out of A, i3 going down out of A : junction
equation for A: i1=i2+i3
Left loop eqn.: + 9 V – i1 (2 Ω) – i3 (1 Ω) – i1(7 Ω) = 0 =>
(leaving off Ω symbol after the numbers, remember that all i’s are multiplied by a
resistance) 9 V = i1(9) + i3 (1) eqn. 1
Right loop eqn., start at A and go CW: -i2(6 Ω) – 6 V + i3(1 Ω) = 0 =>6 V=i3 – 6i2 eqn. 2
From junction eqn., i3=i1-i2, so eqn. 1 becomes 9 V = 10 i1 – i2 eqn. 1’
Eqn. 2, substituting also for i3, becomes 6 V = i1`- 7 i2
Multiply this eqn. by (-10) to get -60V = -10 i1 +70 i2 add this to eqn. 1’ to get
-51 V = 69 i2, so i2= -51/69 A = -0.739 A, from eqn. 2 this gives i3=+1.565 A
From the loop eqn. this gives i1=+0.826 A.
Note that i2 is negative. The current in the right hand loop is counterclockwise.
Using the left loop eqn, the voltage at A is just 9 V – i1(2 Ω) = 7.348 V
The voltage at B is just 0 + i1(7 Ω) = 5.782 V, to VA – VB = 1.57 V
Another way to get the potential difference VA-VB is to start at B and go to A using the
usual rules. Passing through the middle this means that VA-VB = 1× i3 = 1.565 V. Going
through the right branch, we have, VA-VB = 6 + 6i2 = 6×(1- 17/23) = 1.565V. Finally you
could have chosen to go from B to A through the left arm, involving only i1. You would
again get the same answer and come to the conclusion that the potential difference between
two points in a circuit is independent of which path you take and where you “ground” it.
4. A 6 V battery charges a capacitor through a 100 Ω resistor. The capacitor in the circuit
gives a charging time constant of 9 seconds. Assuming that the capacitor starts the cycle
with no charge, after how much time, in seconds, after the switch to the 6 V battery is
closed is the charge on the capacitor = 0.5 Coulombs?
a. 23.4
b. 13.4
c. 18.4
d. 15.9
e. 20.9
Soln.: τ=RC = 100 * C = 9 s => C=(9/100) Farads = 90 mF
Qmax = CV = 0.09*6=0.54 Coulombs
q(t) = Qmax*(1-e-t/RC) so 0.5 Coulombs=0.54 Coulombs*(1-e-t/9)
e-t/9 = 1-0.5/0.54 = 0.07407, take the natural log of both sides, get t/9 = 2.6027, t=23.4 s
5. A magnetic field B=0.1 T directed into the paper at ‘X’ exerts a magnetic force FB on
the positive charge +q=1.6 * 10-19 C as shown in the figure. This centripetal force keeps the
charge rotating in a circular orbit, r=10-6 m. What is the magnitude of the velocity of the
positive charge, in m/s? (mcharge=9.11 * 10-31 kg)
a. 1.8 104
b. 1.8 10-5
c. 1.8 106
d. 1.8 103
e. 1.8 107
Soln.: FB=qvBsinΘ
Θ, sinΘ
Θ=1 This B field force has to supply the centripetal force mv2/r.
mv2/r = qvB
v=qrB/m = 1.6 10-19 * 10-6 0.1 / 9.11 10-31 in units of m/s = 1.8 104 m/s
6. (Student Problem) There is a 1 m long section of a conductor aligned along the y-axis
with mass = 16 10-3 kg in a uniform 0.7 T magnetic field that is pointing to the left in the -x
axis direction. This x-y plane is in the plane of the test paper, and gravity is directed into
the paper. What current (units of Amps) has to flow in the wire and in which direction (+
or - y-direction) for the magnetic force to be up 'out of the paper' on the wire to equal the
downward 'into the paper' force of gravity (the 'floating wire')?
a. 0.22, + y
b. 0.022, - y
c. 0.022, + y d. 0.22, - y
e. 0.002, - y
solution: mg=16 10**-3 kg * 9.8 m/s**2 = 0.1568 N
= F on wire due to B field = BILsintheta = 0.7 T * I * 1 m *sin90
=> I=0.224 A if in +y direction, then force is out of page.
7.
(Student Problem) A current of 1 Ampere flows about two loops of wire (loops A and B –
the loops are pictured together but are far apart and do not interact) in the counter
clockwise direction. The large half circle in each loop, centered at point P marked by the
black dot, has a radius of 10 cm, while the small half circle in each loop (also centered at
point P) has a radius of 5 cm. Calculate the magnetic field produced by the current at
point P in both loops. What is the ratio: (field at point P in loop A/field at point P in loop
B)?
a. 1/3
b. 3
c. ½
d. 2
e. 1
Soln.: B of a loop with N turns is =Nµ
µ0I/2R For our situation, when we calc. the B field
from the 10 cm radius loop, N=1/2 loop, also for the 5 cm half loop. I=1 A.
For the left, loop A: B from 10 cm half loop is out of the paper at P, = ½ µ0 1 A/2*0.1 m
B from the 5 cm half loop is into the paper at P, = ½ µ0 1 A/2*0.05 m
These sum to Bloop Aat P = ½ µ0 1 A/2*0.1 m into the paper – the 5 cm loop dominates because
its radius is smaller.
For the right, loop B: B from 10 cm half loop is out of the paper at P, = ½ µ0 1 A/2*0.1 m
B from the 5 cm half loop is out of the paper at P, = ½ µ0 1 A/2*0.05 m
These sum to Bloop Bat P = 3*[½ µ0 1 A/2*0.1 m] into the paper – the 5 cm loop field adds to
the field from the 10 cm loop. Thus, the B field at P in loop A is 1/3 that in loop B.
8. How many turns per meter should a solenoid have in order to produce a 1 T magnetic
field with 100 A of current?
a. 8000
b. 800
Soln.: B=µ
µ0nI
c. 80
d. 8
e. 80,000
1 T=4π
π 10-7 Tm/A *100A * n
Or n=1/(4π
π 10-5 m) =0.0796 105/m ~ 8000/m
9. A 100 turn coil is arranged so a uniform 2 T external field is perpendicular to the coil
(see sketch.) If the circular coil has r=0.1 m, what is the absolute magnitude of V of the
average induced voltage if the coil is turned so that it is parallel to the field after 0.01 s?
a. 630
b. 310
c. 6300
d. 3100
e. 31
Soln.: Faraday’s law ξ=-N ∆Φ/∆
∆t = 100 2 T (π
π (0.1)2) m2/0.01 s = 630 V
10. A square with one loop of wire with area 1 m2 is moving through space with a constant
velocity of 100 m/s in a region of zero magnetic field. The loop suddenly enters a region of
uniform magnetic field, B=3 T, that is perpendicular to the plane of the square loop.
Between when the loop begins to enter the field region, and 0.01 s later, what is the
magnitude of the average voltage induced in the loop, in V?
a. 300
b. 30
c. 3
d. 3000
e. 30,000
Soln.: |ξ
ξ|=∆
∆ΦB/∆
∆t = 3 T * 1 m2/0.01 s (since the loop moves 1 m in 0.01 s, so the loop starts
empty of field and after 0.01 s is fully in the field region)
So |ξ
ξ|=300 V
11. In the circuit shown in the figure, L = 56 mH, R = 4.6 Ω
and V = 12.0 V. The switch S has been open for a long time
then is suddenly closed at t = 0. At what value of t (in msec)
will the current in the inductor reach (1.1/2.1/0.57) A?
Answer: (6.67/19.9/3.0)
Soln.: The current rises as i(t) = Io (1-e(-t/τ)) where Io = V/R and
τ = L/R. For the first case (i(t) = 1.1 A), we have t = - τ ln[1-i(t)/Io] = [56 ln(11.1/12/4.60)]/4.6 = 6.67 ms.
12. In the previous problem, what is the total energy stored in the inductor a long time
after the switch is closed?
Soln.: After a long time, i.e. once the current has been established, the current is given by
V/R = 12/4.6 = 2.61 A. The energy stored in an inductor is U
= ½LI2 = ½ ×56×10-3(12/4.6)2 = 0.19J. This is also the energy
of the magnetic field.
13. A conducting bar can slide with no friction along two
conducting rails separated by distance L = 0.5 m, as shown
in the figure. A uniform 2 Tesla magnetic field points into
the page, as indicated by crosses, and the resistance R = 0.5
Ω. An external force Fapp pulls the bar to the right with a constant speed v = (3/2/1) m/s.
What is the power (in Watts) provided by the external force Fapp?
Answer: (18/8/2) W
Soln.: The bar is moving at a uniform velocity and therefore must have no force on it.
There are two forces acting on it: the external force Fapp and the induced force Find. These
forces have the opposite direction but must be equal in magnitude. At speed v, the induced
force is (BL)2v/R = |Fapp| = 6 N. The power provided by the external source Fv = 18W.
14. In the previous problem, what is the magnitude of the external force Fapp (in N)?
Answer: (6/4/2) N
The answer is derived in the previous problem.
15. The remnants of a red giant star form a ring around a black hole of mass M= (2/4/6) x
1032 kg, just like the rings of rock and ice around Saturn. Our home planet lies in the plane
of rotation of the ring; the material on one side moves towards us, the material on the other
side moves away from us. The diameter of the ring is D= 2.4x1013 m. The Hubble space
telescope observes the emission from hydrogen atoms inside the ring. Hydrogen atoms at
rest emit light at 410 nm. What is the frequency difference (in 109Hz = GHz) between the
light emitted by Hydrogen atoms flying away from us and the light emitted by Hydrogen
atoms flying towards us in the ring? Assume that the relative velocity between the black
hole and earth can be neglected. Hint: Recall that the gravitational force between two
masses is FG = Gm1m2/r2.
Soln.: This is a Doppler effect problem. The difference in the two frequencies is ∆f = 2u/λ,
where λ is the wavelength of the light emitted (410 nm). To calculate the speed u, assume
that the ring rotates in a circle of radius r = D/2. In a circular orbit, mu2/r = GmM/r2 or u
= √[2GM/D]. Thus ∆f = 1.6264 × 1011 Hz.
16. A small spherical particle with diameter 0.2mm and density 2 g/cm3 is suspended over a
table by a vertical laser beam. Assume that the particle is totally absorbing. What is the
intensity of the laser beam in W/m2?
Soln.: On p.719 of the textbook, example 21.9, there is an expression for the force exerted
by a light beam, F = IA/c where I is the intensity of the light beam, A is the beam area and c
is the speed of light. The expression is for the case when the light beam covers the particle
and is fully absorbed. Since the gravitational force is ρ(4πr3/3)g = I × πr2/c , I = (4/3)ρrgc =
7.848 × 108 W/m2.
17. An object is located 20 cm from a mirror that forms a real image that is twice the size
of the object. The type of mirror and its radius of curvature (in cm) must be, respectively:
Soln.: m = -q/p. The image is said to be real, which means that q >0. Therefore m = - 2 or
q = 2p and using 1/p+1/q = 1/f, f = 40/3. The mirror is concave (f>0) and its radius of
curvature R = 2f = 80/3 cm.
18. A small underwater pool light is located at the bottom of a swimming pool. The circle
of light on the surface of the pool, through which the light comes out, has a radius of
(1.13/1.67/2.27 m. What is the depth (in meters) of the pool? (nwater = 1.333)
Answer: (1/1.5/2)
Soln.: The light is hitting the surface at the critical angle θ = arcsin (1/n). The depth D of
the pool is related to the radius r of the circle above so that r = D tanθ. D = r cotan θ.
19. A man standing 1.52 m in front of a shaving mirror produces an inverted image 18 cm
in front of it. How close (in cm) to the mirror should he stand if he wants to form an
upright image of his chin that is (twice/three times/four times) the chin's actual size?
Answer: (8.05/10.73/12.07)
Soln.: p = 152 cm and q = 18. The mirror equation (as in problem 17 above) gives f = 16.09
cm. Since the desired m (2, 3, 4 all >0), q = -mp and solving from 1/p-1/(mp) = 1/f, we get
the answers for p.
20. A narrow beam of ultrasonic waves reflects off the live
tumor as shown in the figure. If the speed of the wave is
10% less in the liver than in the surrounding medium,
determine the depth of the tumor in cm.
Soln.: There are two ideas that come in to play here. One
is
that we can use the concepts developed for em waves, for
sound waves and secondly that the index of refraction is
inversely proportional to the speed of the excitation involved, i.e. n2/n1 = v1/v2 where the v’s
are the velocities in the corresponding medium.
The depth D = r cotanθ and n1sin50 = n2 sin θ. Or sinθ = (v2/v1) sin50 and v2/v1= 0.9. The
answer follows, D = 6.3 cm.