107
First PHYSICS Midterm Exam
SOLUTIONS
8 February 2001
Last Name (print) First Name (print) Signature SS # Instructor
Problem I. II. III. IV. Total
Max. Points 25 25 25 25 100
Score
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Any other violation of academic honesty can have the same effect. Perform the necessary calculations in the spaces provided under the questions and on the back of the question sheets if additional space is needed. ALL WORK MUST BE CLEARLY SHOWN TO RECEIVE FULL CREDIT.
I. (25 pts) A 10-turn rectangular loop of wire with dimensions shown in the figure is situated so that the left edge is at the edge of a region of uniform magnetic field, B = 0.377 B T, at time t = 0. The total resistance R of the loop is 0.860 . The loop is pushed into the field (from the right) at a constant velocity of 1.50 m/s. m (a) (5 pts) Sketch a graph of the flux 19.2 T-m2 through the loop as a function of time. m, max = NBA = (10)(0.377 T)(3.0 m)(1.7 m)
R
E
3.00 m
1.70 m
t 1s 2s 3s
s
(b) (5 pts) Use Faraday's Law to find the induced emf. E = m / t = (19.2 T-m2) / (2.0 s) = 9.61 V
s
(c) (3 pts) Use Lenz's Law to determine the direction of the induced current in the loop. It would be counterclockwise, because flux into the paper is increasing, so the loop tries to counteract this by producing flux out of the paper. This requires a CCW current by the righthand rule. (d) (3 pts) What is value of the current flowing in the loop? I = E / R = (9.61 V) / (0.860 ) = 11.2A (e) (4 pts) What is the force required to push this 10-turn loop into the field? F = N I l B = (10) (11.2 A) (1.70 m) (0.377 T) = 71.8 N or F v = I R, which gives F = I R / v = (11.2 A) (0.860 ) / (1.5 m/s) = 71.8 N (f) (5 pts) How much work is required to push the loop all the way in, through the full 3 m? W = F (x) = (71.8 N) (3.00 m) = 216 J
2 2 2
2/8/01 Physics 107
MIDTERM EXAM #1
Page 2 of 6
N2 II. (25 pts) A long thin solenoid of length 1 m and cross2 sectional area 40 cm contains N1 = 800 closely and uniformly spaced turns of wire. Wrapped tightly around it is an insulated coil of N2 = 3 turns. Assume all the flux from the solenoid passes through the coil. (a) (8 pts) If a current of 5 A flows through the solenoid, N1, what is the magnetic field inside?
N1
l
B = o
NI 800 (5 A) = (4 x 10 - 7 T - m/A) = 5.03 x 10 -3 T l 1m
B = 5.03 x 10-3 T = 50.3 gauss
(b) (6 pts) If the current in the solenoid, N1, increases from 5 A to 10 A in 5 seconds, what is the induced emf in the coil, N2? The flux through coil #2 is 2 = B 1 N2 Asolenoid = (o N1 / l )(N2 Asolenoid ) I1 E2 = 2 / t. = (o N1 N2 Asolenoid / l ) (I1 / t) = [(4 x 10 T-m/A) (800)(3)(40 x 10 m ) / (1 m)](10 A 5 A) / 5 s
-7 -4 2
E 2 = 1.21 x 10-5 V
or, 2 = (5.03 x 10 T) (3) (40 x 10 m ) = 6.03 x 10 T-m , and it doubles over the 5 s, because the current does and B 1 does. Thus = 6.03 x 10 T-m , and 2 / t = 6.03 x 10 T-m / 5 s = 1.21 x 10
-3 2 5 -5 2 -3 -4 2 -5 2
V
(c) (6 pts) Calculate the mutual inductance between the coil and the solenoid. 2 =
B1 (a) =
S
B1 (N 2 Asolenoid ) = (5.03 x 10 -3 T) (3) (40 x 10 -4 m 2 )
= 6.03 x 10 - 5 T - m 2 2 = L2 I2 + M21 I1 Therefore, M21 = 2 / I1 = (6.03 x 10 T-m ) / (5 A)
-5 2
M 21 = 1.21 x 10-5 H = 12.1 H
(d) (5 pts) If the current through the coil, N2, is increased from 0 to 10 A in 5 seconds, what is the induced emf in the solenoid, N1? We have 1 = L1 I1 + M12 I2 = 0 + M12 I2. But M12 = M21. So, -5 5 1 / t = M21 (I2 / t) = (1.21 x 10 H)(10 A / 5 s) = 2.41 x 10 V 2/8/01 Physics 107 MIDTERM EXAM #1 Page 3 of 6
III. (25 pts) An orbiting satellite 150 km above the Earth's surface emits a microwave signal uniformly in all directions at a frequency of 5.00 x 109 Hz with a power of 300 W. (a) (4 pts) Find the wavelength of the EM wave. f = c = c / f = (3.00 x 10 m/s) / (5.00 x 10 s ) = 0.0600 = m 6.00 cm
8 9 1
= 6.00 cm
(b) (6 pts) Find the intensity of the signal at the Earth's surface in W/m . S = P / (4 R ) = (300 W) / [12.57 x (1.50 x 10 m) ]
2 5 2 2
R E Earth's surface
S = 1.06 x 10 9 W/m2
(c) (7 pts) Find the rms values of the electric and magnetic fields.
S = c o Erms2
Erms = [(1.06 x 10
9
Erms = [S / (c o)]
2 8
1/2
Brms = Erms / c
12
W/m ) / (3.00 x 10 m/s x 8.85 x 10
4 8
C /N-m )]
2
2 1/2 12
= 6.32 x 10
4
V/m
Brms = Erms / c = (6.32 x 10
V/m) / (3.00 x 10 m/s) = 2.11 x 10
T
(d) (4 pts) Assuming that the satellite is directly overhead and you observe at one instant an E-field pointing north, along what line is the satellite's linear antenna oriented? Explain your answer.
The antenna must be aligned N-S, because a linear antenna radiates an electric field which oscillates parallel to the wire in the plane to the antenna.
(e) (4 pts) Draw the B-field vector and the wave's velocity vector at this instant. The v must be straight down. The E when rotated into the B with the right hand must have the thumb pointing along v. So, in this figure, B must be out of the paper at this instant.
B E v
Earth's surface
2/8/01 Physics 107
MIDTERM EXAM #1
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IV. (25 pts) Consider two thin converging lenses of focal lengths f1 = 25 cm and f2 = 20 cm. If an object is placed 50 cm to the left of lens #1, the final image produced by this pair of lenses is enlarged by a factor of 2.00 and is upright.
25 cm 50 cm
f1
f2
l
20 cm
(a) (10 pts) Find the image of the object that lens #1 produces by itself.
1 1 1 + = d o1 d i1 f1 Thus,
1 1 1 1 1 1 = = = d i1 f1 d o1 25 cm 50 cm 50 cm
d i1 = 50 cm. Also, m1 = di1 / do1 = (50 cm) / (50 cm) = -1.00
The image is inverted, unchanged in size, and 50 cm to the right of the first lens.
(b) (15 pts) Find the separation distance l between the lenses. (Note: Depending on your answer for l , the focal points between the lenses could overlap instead of as they are shown.) The new object distance is do2 = l d i1 = l 50 cm. If l turns out less than 50 cm, then the new object will be virtual with a negative object distance. m1 m2 = m = + 2.00 = ( 1.00) ( di2 / do2) = + d i2 / (l 50 cm). In fact, m2 = 1.00 m = 2.00 = d i2 / do2 di2 = + 2.00 (l 50 cm)
1 1 1 + = do2 di2 f2
1 l 50 cm
+
1 1 = ( + 2.00)( l 50 cm ) 20 c...