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(cah3459) howard ohw19 turner (56705) This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A wire carries a current of I = 44 A along the x-axis from x1 = 5 cm to x2 = 1.6 cm. so the resultant magnetic eld is B=
1
y
= 126.785 T .
0 I (cos 1 cos 2 ) 4d 0 (44 A) = (cos 38.6598 4 (0.04 m) cos 111.801 )
P x x1 I x2
Find the magnitude of the resulting magnetic eld at the point r = 4 cm on the y axis. Correct answer: 126.785 T. Explanation: Let : x1 = 5 cm , x2 = 1.6 cm , and r = 4 cm = 0.04 m .
002 (part 1 of 2) 10.0 points Imagine a very long, uniform wire that has a linear mass density of 0.0033 kg/m and that encircles the Earth at its equator. Assume the Earths magnetic dipole moment is aligned with the Earths rotational axis. The Earths magnetic eld is cylindrically symmetric (like an ideal bar magnetic). The acceleration of gravity is 9.8 m/s2 and the magnetic eld of the earth is 9 105 T . What is the magnitude of the current in the wire that keeps it levitated just above the ground? Correct answer: 359.333 A. Explanation: Let : g = 9.8 m/s2 , BEarth = 9 105 T , and = 0.0033 kg/m . The Earths geographical North pole is a magnetic South pole and vice versa. The North pole of a compass is attracted to the South magnetic pole of the Earth, the place where Santa Claus lives. The magnetic and gravitational forces must balance. Therefore, it is necessary to have (for unit length of wire) F = I B = g I= g B (0.0033 kg/m) (9.8 m/s2 ) = 9 105 T = 359.333 A .
From the Biot-Savart Law 0 I ds r dB = , 2 4 r the magnetic eld at a point from a straight line segment is B= In this case, 1 = arctan y x1 4 cm = arctan (5 cm) = 38.6598 , and y 2 = 180 arctan x2 4 cm = 180 arctan 1.6 cm = 111.801 , 0 I (cos 1 cos 2 ) . 4r
howard (cah3459) ohw19 turner (56705) 003 (part 2 of 2) 10.0 points The current in the wire goes in the 1. same direction as the Earths spinning motion (West to East). correct 2. opposite direction as the Earths spinning motion (East to West). Explanation: Since the Sun rises in the East, the Earth is turning towards the East and the Earths angular velocity is parallel to the Earths axis, pointing from South to North.
S
2
Let : r = 0.8 m , I0 = 2.4 nA = 2.4 109 A = 0.006 m . The electric eld is E = J I0 = A I0 = r2 = (0.006 m) 2.4 109 A (0.8 m)2
and
= 7.16197 1012 . 005 (part 2 of 2) 10.0 points The current in the wire changes with time according to I = I0 + c t, where I0 is as above and c = 7 105 A/s. Find the total magnetic eld a distance 3.7 m from the axis at time 2 1015 s . Correct answer: 4.73919 1014 T. Explanation:
vE
N
B
vEarth
FI The total force on the wire is zero and the gravitational force is downward (inward), so the magnetic force must be upward (outward). If the current goes from West to East, in the same direction as the Earths spinning motion, I B will be vertically upward (outward). 004 (part 1 of 2) 10.0 points A long, solid cylindrical conductor of radius 0.8 m carries a current of 2.4 nA parallel to its axis and uniformly distributed over its crosssection. The conductor has a resistivity of 0.006 m. Find the value of the electric eld inside the conductor. Correct answer: 7.16197 1012 V/m. Explanation:
Let : c = 7 105 A/s , r = 3.7 m and t = 2 fs = 2 1015 s . Ienc = J A = I and B ds = 0 Ienc + 0 so B 2 r = 0 I r2 + 0 b2
0 0
r2 b2 d E dt r2 d I b2 d t
B=
dI 0 r I + 0 2 b2 dt 0 r (I0 + c t + 0 c) = 2 b2
howard (cah3459) ohw19 turner (56705) = 0 (3.7 m) 2.4 109 A 2 (0.8 m)2 + (7 105 A/s) (2 1015 s) + (8.85419 1012 C2 /N m2 ) 007 10.0 points The wire is carrying a current I . y 180
3
(0.006 m)(7 105 A/s) I
= 4.73919 1014 T . 006 10.0 points A long straight wire lies on a horizontal table and carries a current of 1.41 A. A proton moves parallel to the wire (opposite the current) with a constant velocity of 11200 m/s at a distance d above the wire. The acceleration of gravity is 9.8 m/s2 . Determine this distance of d. You may ignore the magnetic eld due to the Earth. Correct answer: 3.08777 cm. Explanation: Let : qp v I mp = 1.6 1019 Coulomb , = 11200 m/s , = 1.41 A , = 1.67 1027 kg , and
I r
x I Find the magnitude of the magnetic eld B at O due to a current-carrying wire shown in the gure, where the semicircle has radius r , and the straight parts to the left and to the right extend to innity. 1. B = 2. B = 3. B = 4. B = 5. B = 6. B = 7. B = 8. B = 0 I 2r 0 I 4r 0 I 2r 0 I correct 4r 0 I 3r 0 I 3r 0 I r 0 I r
O
0 = 1.25664 106 T m/A . Let the current I ow to the right; then it creates a magnetic eld 0 I B= 2d at the protons location This magnetic eld induces an upward force on the proton which balances the weight, so we have mp g = qp v 0 I 2d qp v 0 I d= 2 mp g (1.6 1019 Coulomb) (11200 m/s) d= 2 (1.67 1027 kg) (9.8 m/s2 ) (1.25664 106 T m/A) (1.41 A) = 3.08777 cm .
Explanation: By the Biot-Savart Law, B= 0 I 4 ds r . 2 r
Consider the left straight part of the wire. The line element ds at this part, if we come in from , points towards O, i.e., in the xdirection. We need to nd d s to use the r Biot-Savart Law. However, in this part of the wire, is pointing towards O as well, so d s r
howard (cah3459) ohw19 turner (56705) and are parallel meaning d s = 0 for this r r part of the wire. It is now easy to see that the right part, having a d s antiparallel to , also r gives no contribution to B at O. Let us go through the semicircle C. The element d s, which is along the wire, will now be perpendicular to r, which is pointing along the radius towards O. Therefore | d s | = ds r using the fact that r is a unit vector. So the Biot-Savart Law gives for the magnitude B of the magnetic eld at O: 0 I B= 4 ds . r2 A I 2R The Biot-Savart Law is given by dB = r 0 I d . 2 4 r R Explanation: Let : 0 = 1.25664 106 T m/A I = 2.52 A and R = 7.08 cm .
4
C
Since the distance r to the element ds is constant everywhere on the semicircle C, we will be able to pull it out of the integral. The integral is ds 1 =2 2 r r ds =
C
R
(1)
C
1 LC r2
where LC = r is the length of the semicircle. Thus the magnitude of the magnetic eld is B= 0 I 0 I 1 . r = 2 4 r 4r
Note: The distance from a current element on a circle to the center is a constant, namely r , so we can pull this out of the integral. Also, the current element, I dl, is always perpendicular to , so sin = 1. r Hence, Bf ull circle = 0 I d 4 r2 I = 0 2 2r 4 r 0 I = , 2r
008 (part 1 of 2) 10.0 points The loop in the gure carries a current 2.52 A. The semicircular portion has a radius 7.08 cm. The permeability of free space is 1.25664 106 T m/A .
8c m
A 2.52 A 2 7.08 cm
7.08 cm
where a half-circular arc is one-half this value. Assume: Out of the page to be positive. The angle (between the vector and the vector r ) is from the Biot-Savart Law, Eq. (1), for the magnetic eld B a distance a away from a wire segment 0 I 2 B= sin d 4 a 1 I = 0 (cos 1 cos 2 ) , 4a where a = R for the sides of the semi-square in gure.
Determine the magnitude of the magnetic eld at A. Correct answer: 2.12492 10
5
7. 0
T.
howard (cah3459) ohw19 turner (56705) is B= 2 r 3 4 ll 0 I (cos 1 cos 2 ) 4R I 1 1 =0 4R 2 2 I 2. =0 4R
5
R
r lr 3 4 lr 4 2 r
ll
R
Then the magnetic eld at the center is Bsquare = 4 B 2 0 I = . R The magnetic eld at the center of a circular loop is Bcircle = 0 I . 2R
lb 4
r
The magnetic eld due to the semicircular portion and the straight wire segments is B= 1 I 0 semicircle 2 2R I cos cos +0 4R 2 4 0 I 3 + cos cos 4R 4 4 I 3 +0 cos cos 4R 4 2 2 I 1 + = 0 R 4 2 =
r ight edge
Therefore the magnetic at eld the center of the half-circular loop plus half-square loop is B= 1 [Bcircle + Bsquare] 2 1 0 I 2 0 I = + 2 2R R I 1 2 =0 + , R 4 2
bottom
lef t edge
(2)
B = 2.12492 105 T ,
(1.25664 106 T m/A) (2.52 A) 0.0708 m 0.475079 = 2.12492 105 T
which is the same as Eq. (2). 009 (part 2 of 2) 10.0 points In which direction is the magnetic eld pointing at A ? 1. not enough information given 2. into the page correct 3. out of the page
where B is pointed into the page. Alternate Solution: Consider a square loop with each side a length 2 R.
1
R
2
For the line segment (one-fourth of a square), the magnitude of the magnetic eld
Explanation: The current is circulating in the clockwise direction. Using the right hand rule, with the thumb pointing in the direction of the current, our ngers projected unto point A is into the page.
howard (cah3459) ohw19 turner (56705) Thus, we nd 010 (part 1 of 2) 10.0 points The magnetic coils of a tokamak fusion reactor are in the shape of a toroid having an inner radius of 1.067 m and outer radius of 1.311 m. The toroid has 1171 turns of wire, each of which carries a current of 11.48 kA. B = 2.26124 T .
6
011 (part 2 of 2) 10.0 points What is the the magnetic eld strength a distance b + 2 a from the center of the tokamak? 1. B = 2 0 N I (b + 2 a) 4N I 2. B = 0 (b + 2 a) 3 0 N I 3. B = (b + 2 a) 0 N I 4. B = (b + 2 a) 5. B = 0 correct 6. B = 7. B = 0 N I (b + 2 a) 0 N I 4 (b + 2 a) 2N I 0 (b + 2 a) 0 N I 2 (b + 2 a) 0 N I 3 (b + 2 a)
a r
b
Find the magnitude of the magnetic eld at a+b , from the center of the a distance r = 2 tokamak. Correct answer: 2.26124 T. Explanation: Let : N r I 0 = 1171, = 1.189 m, = 11.48 kA, and = 1.25664 106 T m/A.
8. B = 9. B = 10. B =
Here currents are coming out of the paper on the inner radius and returning into the paper on the outer radius. By Ampres law, e B d = 2 r B = 0 Itotal Itotal = N I . Therefore, B= 0 N I 2r (1.25664 106 T m/A)(1171)(11.48 kA) = 2 (1.189 m) = 2.26124 T .
Explanation: An Amperian loop that extends just beyond the outer radius of the toroid will include every piece of wire in the diagram. Since, for every coil on the inside, there is a corresponding coil on the outside, and since the coils on the inside and outside have current owing out of and into the page, this Amperian loop has no net current piercing its surface. Therefore, there is no net magnetic eld anywhere outside of the outer radius. Thus B = 0 . 012 (part 1 of 2) 10.0 points Consider two long parallel wires which are perpendicular to the plane of the paper (i.e., the x-y plane). Both wires carry the same current, I .
For r we take the average of the value of the inner radius a and the outer radius b. Convert A . current to amps by multiplying by 103 kA
howard (cah3459) ohw19 turner (56705) Wire #1 intersects the plane a distance a above point O and wire #2 intersects the plane a distance a below point O. Point C is equidistant from both wires and is a distance a from point O. a a 45 C
7 wire #1
wire #1 II O a III O
y I x IV
B2
C
B1
O wire #2
45
wire #2 a D What is the direction of the magnetic eld at C ? 1. in quadrant IV 2. in the positive y direction 3. in the negative y direction correct 4. in quadrant I 5. in the positive x direction 6. in quadrant II 7. in the negative x direction 8. into the plane 9. in quadrant III 10. out of the plane
From this gure, we can see by symmetry that the y components of the magnetic elds cancel, leaving only the component in the positive x direction. 013 (part 2 of 2) 10.0 points What is the magnitude of the magnetic eld at C ? 1 0 I 1. BC = 2 2 a 1 0 I 2. BC = 4 a 1 0 I 3. BC = correct 2 a 1 0 I 4. BC = 2a 5. BC = 0 1 0 I 6. BC = 2 a 1 0 I 7. BC = 4a 1 0 I 8. BC = 22a 1 0 I 9. BC = 2a 1 0 I 10. BC = a Explanation: Note: The distance r from each wire to C is a = 2a. r= sin 45
Explanation:
The magnitude of the x component of the magnetic eld due to each wire at C is given
howard (cah3459) ohw19 turner (56705) by 0 I B1 = B2 = cos 45 2 2 a 1 0 I = 2 2 a 2 0 I = . 4a Hence the contribution from both wires at C is just twice this value: BC = 2 B1 = 0 I . 2a
L
8
rb ra
The current I = 2 mA . Since the cylindrical shell is innitely long, and has cylindrical symmetry, Amperes Law gives the easiest solution. Consider a circle of radius r1 centered around the center of the shell. To use Amperes law we need the amount of current that cuts through this circle of radius r1 . To get this, we rst need to compute the current density, for the current owing through the shell. J= = I A
2 rb
014 10.0 points A long cylindrical shell has a uniform current density. The total current owing through the shell is 2 mA. The permeability of free space is 1.25664 106 T m/A .
11 k m
7 cm 3 cm
I 2 ra (2 mA) = [(0.07 m)2 (0.03 m)2 ] = 0.159155 A/m2 . The current enclosed within the circle is
The current is 2 mA . Find the magnitude of the magnetic eld at a point r1 = 4.1 cm from the cylindrical axis. Correct answer: 1.90488 nT. Explanation:
2 2 Ienc = [r1 ra ] J = [(0.041 m)2 (0.03 m)2 ] (0.159155 A/m2 ) = 0.0003905 A .
Amperes Law, B ds = 0 Ienc B 2 r1 = 0 Ienc 0 Ienc B= 2 r1 1.25664 106 T m/A = 2 (0.041 m) (0.0003905 A) = 1.90488 nT .
Let : L = 11 km , ra = 3 cm = 0.03 m , rb = 7 cm = 0.07 m , r1 = 4.1 cm = 0.041 m , I = 2 mA , and b = 1.25664 106 T m/A .
howard (cah3459) ohw19 turner (56705) 015 (part 1 of 3) 10.0 points A current of 14 A ows into a capacitor having plates with areas of 0.5 m2 . The permittivity of free space is 8.85 1012 C2 /N m2 . What is the displacement current between the plates? Correct answer: 14 A. Explanation: Let : I = 14 A . Because of charge conservation, we have Id = I = 14 A . 016 (part 2 of 3) 10.0 points dE What is between the plates for this curdt rent? Correct answer: 3.16384 1012 V/m s. Explanation: Let : The line integral of B dl is
C
9
B dl = 0 Ienclosed r2 Id A = 4 107 T m/A = 0 (0.2 m)2 (14 A) 0 . 5 m2
= 4.42158 106 T m . 018 10.0 points A circular loop of wire of radius R = 3.83 m lies in the xy -plane, centered about the origin. The loop is carrying a current of I = 6.13 A owing in counterclockwise direction. Consider two l = 1.17 mm segments of the loop: one centered about the positive x-axis, the other centered about the positive y -axis. Hint: Use Biot-Savart law. The permeability of free space is 1.25664 106 Tm/A .
y l2 =ds2
A = 0.5 m2 and 12 2 C / N m2 . 0 = 8.85 10
I2 R r l1 =ds1 I1 x
dE Id = dt 0A = (8.85 1012 14 A C2 /N m2 ) (0.5 m2 )
= 3.16384 1012 V/m s .
017 (part 3 of 3) 10.0 points What is the line integral of B dl around a circle of radius 20 cm that lies within the plates and parallel to the plates? Correct answer: 4.42158 10 Explanation: Let :
6
What is the magnitude of the force the rst exerts on the second? Correct answer: 1.2398 1013 N. Explanation:
T m. Let : 0 = 1.25664 106 Tm/A , l = 1.17 mm = 0.00117 m , I = 6.13 A , and R = 3.83 m .
r = 20 cm = 0.2 m and 0 = 4 107 T m/A .
howard (cah3459) ohw19 turner (56705) Since the radius, R, of the current loop is much bigger than the length, l, of both small segments, we can regard the segments as a small straight line. From the Biot-Savart law dB = 0 I ds r . 2 4 r
10
The eld B produced by the rst segment at the position of the second is B= 0 I 4 0 I = 4 0 I = 4 ds r 2 r ly r ( 2 R )2 l sin 45 z . 2 2R
The force exerted by the rst segment on the second one then is F = I l2 B = I l(1) x B 2 0 I l2 sin 45 ( ) xz = (1) 2 4 2R 2 l 2 sin 45 y 0 I = 2 4 2R (1.25664 106 Tm/A) (6.13 A)2 = 4 2 sin 45 y (0.00117 m) 2 2 (3.83 m) = 1.2398 1013 N y . So the magnitude of force is 1.2398 1013 N .

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University of Texas - PHY 303L - 56705

nguyen (jmn727) oldhomework 24 Turner (59070) This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A toroid having a rectangular cross section (a = 1.22

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rosen (arr956) Homework 08 Chiu (58295) This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A 3 m long wire weighing 0.087 N/m is suspended directly ab

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homework 09 FAUSAK, TAYLOR Due: Mar 24 2008, 4:00 am Question 1, chap 31, sect 1. part 1 of 2 10 points In the arrangement shown in the gure, the resistor is 5 and a 8 T magnetic eld is directed into the paper. The separation between the rails is 8 m . Ne

University of Texas - PHY 303L - 56705

homework 09 FAUSAK, TAYLOR Due: Mar 24 2008, 4:00 am Question 1, chap 31, sect 1. part 1 of 2 10 points In the arrangement shown in the gure, the resistor is 5 and a 8 T magnetic eld is directed into the paper. The separation between the rails is 8 m . Ne

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hopkins (tlh982) HW07 criss (4908) This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A long cylindrical shell has a uniform current density. The total

University of Texas - PHY 303L - 56705

exam 03 ANDERSON, ZACH Due: Apr 3 2008, 11:00 pm1E & M - Basic Physical ConceptsElectric force and electric eldElectric force between 2 point charges: |F | = k 1r2 2 k = 8.987551787 109 N m2 /C2 1 12 C2 /N m2 0 = 4 k = 8.854187817 10 qp = qe = 1.60217

University of Texas - PHY 303L - 56705

exam 03 ANDERSON, ZACH Due: Apr 3 2008, 11:00 pm1E & M - Basic Physical ConceptsElectric force and electric eldElectric force between 2 point charges: |F | = k 1r2 2 k = 8.987551787 109 N m2 /C2 1 12 C2 /N m2 0 = 4 k = 8.854187817 10 qp = qe = 1.60217

University of Texas - PHY 303L - 56705

garza (olg97) Homework No. 6 jimenez (1302-01) This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points The magnetic eld of the Earth at a cer

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Platt, David Quiz 2 Due: Oct 18 2005, 10:00 pm Inst: Ken Shih This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 po

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Shie, Gary Homework 26 Due: Nov 5 2004, 4:00 am Inst: Turner This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 poi

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Kapoor (mk9499) oldhomework 27 Turner (60230) This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points An AC power supply produces a maximum v

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Kapoor (mk9499) oldmidterm 01 Turner (60230) This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points Two spheres, fastened to pucks, are riding on a fricti

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johnson (rj6247) hw 13 Opyrchal (121014) This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points An inductor has a 66.3 reactance at 91.2 Hz. What will be

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howard (cah3459) hw 23 turner (56705) This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points The alternating voltage of a generator is repre

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nguyen (jmn727) homework 27 Turner (59070) This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 3) 10.0 points An AC voltage of the form V = Vmax sin 2 f

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terry (ect328) homework 32 Turner (59130) This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 3) 10.0 points In an LR circuit, a 174 V(rms), 78 Hz sourc

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Version One Homework 4 Juyang Huang 24018 Mar 28, 2008 This print-out should have 43 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. Electromagnetic Wave 01 34:02,

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howard (cah3459) hw24 turner (56705) This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 6) 10.0 points You are designing a radio receiver based on the

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Platt, David Homework 26 Due: Nov 9 2005, 4:00 am Inst: Ken Shih This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 3) 10

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Platt, David Homework 26 Due: Nov 9 2005, 4:00 am Inst: Ken Shih This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 3) 10

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howard (cah3459) ohw23 turner (56705) This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points When a particular inductor is connected to a si

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howard (cah3459) ohw24 turner (56705) This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 7) 10.0 points Consider a parallel circuit with a 99 resistor,

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Create assignment, 53823, Homework 12, Dec 05 at 2:54 pm This print-out should have 17 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: nd all choices before making your selection. The due time is Ce

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Patel, Kinal Homework 17 Due: Apr 4 2008, 7:00 pm Inst: Weathers This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. 001 (part 1 of 1) 10

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terry (ect328) homework 15 Turner (59130) This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points Note: Making use of the fact that the resistors and elect

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terry (ect328) homework 16 Turner (59130) This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A 0.03 F capacitor is given a charge Q0 . After 9 s, the

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terry (ect328) oldhomework 14 Turner (59130) This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points A 3.2 g wire has a density of 13.5 g/cm3

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terry (ect328) oldmidterm 02 Turner (59130) This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A capacitor network is shown in the following gure. 4.3

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midterm 02 VARELA, CHRISTOPHER Due: Oct 18 2007, 11:00 pm1E & M - Basic Physical ConceptsElectric force and electric eldElectric force between 2 point charges: |F | = k 1r2 2 k = 8.987551787 109 N m2 /C2 1 12 C2 /N m2 0 = 4 k = 8.854187817 10 qp = qe

University of Texas - PHY 303L - 56705

midterm 03 VARELA, CHRISTOPHER Due: Nov 15 2007, 11:00 pm Question 1, chap 32, sect 5. part 1 of 1 10 points In a series RLC ac circuit, the resistance is 8 , the inductance is 25 mH, and the capacitance is 24 F. The maximum potential is 219 V, and the an

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midterm 02 SAENZ, LORENZO Due: Mar 6 2008, 9:00 pm1E & M - Basic Physical ConceptsElectric force and electric eldElectric force between 2 point charges: |F | = k 1r2 2 k = 8.987551787 109 N m2 /C2 1 12 C2 /N m2 0 = 4 k = 8.854187817 10 qp = qe = 1.602

University of Texas - PHY 303L - 56705

howard (cah3459) extra credit 01 turner (56705) This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points How much positive charge is in 0.5 kg of uorine? The

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howard (cah3459) extra credit 02 turner (56705) This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 2) 10.0 points Consider two concentric spherical con

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howard (cah3459) hw13 turner (56705) Power (W) This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A variable resistor is connected across a constant v

University of Texas - PHY 303L - 56705

howard (cah3459) hw13 turner (56705) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 (part 1 of 3) 10.0 points Consider the circuit 26 V 3. 6 2. 5 I1 a 3. 8 A

University of Texas - PHY 303L - 56705

The Electric FieldStudy Guide for Chapter 23Outline1. The Electric Field of Point Charge The electric field is a vector field associated with a charge distribution. By vector field, we mean that the electric field consists of one vector E for each poin

University of Texas - PHY 303L - 56705

Gauss' LawStudy Guide for Chapter 24Outline1. Electric Flux Electric flux represents the total number of electric field lines crossing through a surface. The basic equation for flux is F IE where I is the magnitude of the electric field, and E is the a

University of Texas - PHY 303L - 56705

The Electrostatic PotentialStudy Guide for Chapter 25Outline1. The Electrostatic Potential Every point in space has an electrostatic potential. Potential is a scalar that measures the potential energy per unit charge that a positive test charge would h

University of Texas - PHY 303L - 56705

CapacitorsStudy Guide for Chapter 26Outline1. Capacitor A capacitor is a device that stores charge. Usually, a capacitor stores an equal amount of positive and negative charge, held close together on two metal pates. The charge U on the capacitor is pr

University of Texas - PHY 303L - 56705

Current and Ohm's LawStudy Guide for Chapter 27Outline1. Electric Current Electric current is the flow of electric charge. The magnitude of the current is the charge per unit time that flows through the conductor: M U ?> or M .U .>Current is measured

University of Texas - PHY 303L - 56705

Direct Current CircuitsStudy Guide for Chapter 28Outline12. Electromotive Force Most direct current circuits are powered by a battery or power supply that produces a constant potential difference between its two ends. Such a potential difference is kno

University of Texas - PHY 303L - 56705

Magnetic Force and FieldStudy Guide for Chapter 29Outline12. Magnetic Fields A magnetic field can exert force on any moving charge. The force F on a charge is F v B where v is the velocity of the charge, and B is the magnetic field vector at the locati

University of Texas - PHY 303L - 56705

Charges and Currents in Magnetic FieldsStudy Guide for Chapter 30Outline1. Circular Motion in a Uniform Magnetic Field A charged particle moving perpendiuclar to a uniform magnetic field will travel in a circle. The radius of this circle is given by th

University of Texas - PHY 303L - 56705

Electromagnetic InductionStudy Guide for Chapter 31Outline1. Motional EMF A straight segment of wire moving through a magnetic field experiences an induced EMF. The magnitude of the EMF is given by the formula where is the length of the wire, is the ve

University of Texas - PHY 303L - 56705

Alternating CurrentStudy Guide for Chapter 32 Oscillations Alternating current involves current that oscillates over time. We usually the following formula for oscillations:Cmax>maxone period max cosHere max is the amplitude, and is the angular fr

University of Texas - PHY 303L - 56705

Induction of Magnetic FieldsStudy Guide for Section 33.1 Just as a changing magnetic field can induce an electric field, a changing electric field can induce a magnetic field. This induced field obeys the following equation: closed B L loop has been su

University of Texas - PHY 303L - 56705

RC and RL CircuitsStudy Guide Exponential Functions In physics, most exponential functions are governed by a time constant which has units of time. The following graph shows a typical exponential decay: (Exponential Decay)Sometimes an exponential funct

Grambling State - CS - 10268

Programming Language Concepts Name: _ Date: _ 1. Write regular expressions to capture the following.Numeric constants in C. These are octal, decimal, or hexadecimal floating-point values. An octal integer begins with 0, and many contain only the digits 0

LSU - AAST - 2001

PART I Lesson Title: Airpower through WWI Teaching Method: Informal Lecture/Guided Discussion/Embedded Video Time Required: 3 hours Prerequisite Classes: None Interrelated Information: All Lessons Instructor Reference: AFDD 1 (Examples of discussed items)

Nova Southeastern University - BUSINESS - 0000

Free response Unit 3

CUNY City Tech - PSY - 1101

Chapter 3: Neurological and Genetic Bases of Behavior 3-1 The Neuron Santiago Ramn y Cajal (1852-1934): discovered that the brain consisted of an untold number of separate, distinct cells. Perhaps millions of separate cells. Further, these specialized cel

University of Phoenix - FINANCE - FIN 486

Huffman Trucking Co.MemoTo: Department Managers From: Huffman Trucking CFO Date: November 22, 2010 Re: Strategic Plan & Budget Plan Each manager will be responsible for planning, implementing, and delegating the strategic plan in their department. The m

Humber - BIO - 140

Chapter 6Cellular Respiration6.1 The Chemical Basis of Cellular Respirationa. Food as fuel a. The principle of redox a. Cellular respiration is controlled combustionEnergy Flow The SunUltimate source of energy Photosynthesis Captures energy of li

CUNY Baruch - COM - 1500

Dinesh Ghising Prof. Anne Hofmann Pub 1250 Topic: Mass transit in NY Oct 15th 2010 Months After Cutbacks, M.T.A. Approves Higher Fares According to New York Times reporter, Michael M. Grynbaum, The Metropolitan Transportation Authority approved a package

CUNY Baruch - COMM - 1010

Philosophy 1500 exam review sheet10/26/10 3:40 PM Descartes: Epistemology is the study of knowledge and, at least for what concerns us so far, the study of what we take to be true and justified beliefs. Hyperbolic doubt

Christopher Newport University - WS - ws320

JOURNAL(1)About COCO CHANELLast week have seen a movie called ,I Coco avant Chanel ,thereasonwhyIwenttoseethismovieisIwas so boring to open the laptop with nothing to do and I pressedthemouseonthePPSwhichisaonlinesoftware aboutwantchingmovie,thenIgotthi

University of Texas - CH 301 - 50985

elfenbein (ee4265) Homework 9 sutclie (50985) This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. Whats the chemical formula of water? HIJKLMNO (a little geek humor fo

UCLA - EE - 110

EE110 Midterm Answer1. (a) Isc = -j100/(100+j200) = -0.4-j0.2 (b) Voc = -j100(j100)/(100+j300) =10-j30 (c) Zn = Voc/Isc= 10+j70 2. Z_AB(s)=5R/(5+3sCR) Z_AB(jw)=5R/(5+j3wCR) 3. I = V*(s^2LC)/(s^2LCR+SL+R) P(dissipated by R) =100W Real power =100 W Reactiv

DeVry Denver - BIOS - biology

Elizabeth ClarkFlylab Lab Notes for DeVry Biology 1.(eye color)Results of Cross Parents(Female: +) x (Male: SE)OffspringPhenotype Female: + Male: + TotalNumber Proportion 4971 4923 9894 0.5024 0.4976Ratio 1.010 1.000-these are the results to my f1

Pontificia Universidad Javeriana - H - h

A mis amigo s!Wunderland Bei Nacht.Bert Kaempfert.Hace muchsimos aos, un joven recin casado estaba sentado en un sof en un da caluroso y hmedo, bebiendo jugo helado durante una visita a su padre.Mientras conversaba sobre la vida, el matrimonio, las re

Marquette - PHIL - 1001

Girolami 1 Sara Girolami Teacher: Dawn DiNicola Philosophy 1001-702 October 22, 2010Free Will: A Necessary EvilWhere is God in a world filled with unspeakable evil? This is a question that has been pondered for years and the answer is still a mystery. E