course PHY 232 These are all of the queries from Assignments 25-31 for Electrity and Magnetism. qDԪԺ~assignment #025 025. Physics II 10-26-2008
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12:12:29 Query introductory set #1, 10-17 Explain how to find the potential difference in volts between two given points on the x axis, due to a given charge at the origin.
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RESPONSE --> Volts are the potential difference (V2-V1). However, if we have work(joules) and charge, we divide Joules by Coulombs and we have volts(the potential difference) confidence assessment: 2
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12:14:03 ** Potential difference is the work per Coulomb of charge moved between the two points. To find this work you can multiply the average force on a Coulomb of charge by the displacement from the first point to the second. You can find an approximate average force by finding the force on a 1 Coulomb test charge at the two points and averaging the two forces. Multiplying this ave force by the displacement gives an approximate potential difference. Since the force is not a linear function of distance from the given charge, if the ratio of the two distances from the test charge is not small the approximation won't be particularly good. The approximation can be improved to any desired level of accuracy by partitioning the displacement between charges into smaller intervals of displacement and calculating the work done over each. The total work required is found by adding up the contributions from all the subintervals. University Physics students should understand how this process yields the exact w ""ork, which is the integral of the force function F(x) = k Q / x^2 between the two x values, yielding total work W = k * Q * 1 Coulomb ( 1 / x1 - 1 / x2) and potential difference V = k * Q ( 1 / x1 - 1 / x2). **
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RESPONSE --> okay. This is not quite what I said, however diving work by charge will result in the potential difference. Work is found by taking the product of force and displacement. self critique assessment: 2
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12:16:28 Explain how to find the potential difference between two points given the magnitude and direction of the uniform electric field between those points.
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RESPONSE --> The Electric Field E is the quotient of Force over charge (Newtons per Coulomb). Apparently a Volt per meter is equivalent to this. If we could then multiply by the displacent 'ds, we would have just the potential difference. confidence assessment:
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12:16:41 ** The work per Coulomb done between the two points is equal to the product of the electric field E and the displacement `dr. Thus for constant field E we have V = E * `dr. **
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RESPONSE --> awesome self critique assessment:
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12:17:22 Explain how to find the average electric field between two points given a specific charge and the work done on the charge by the electric field as the charge moves between the points.
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RESPONSE --> The average strength of an electric field is measured as the average force per unit charge. confidence assessment:
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12:17:48 ** You get ave force from work and distance: F_ave = `dW / `ds. You get ave electric field from work and charge: E_ave = F / q. An alternative: Find potential difference `dV = `dW / q. Ave electric field is Eave = `dV / `ds **
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RESPONSE --> okay. I used the last one at the bottom. self critique assessment:
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12:18:59 In your own words explain the meaning of voltage.
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RESPONSE --> Voltage is the strength of a field. It does not have a flow,(that would be current), but it state how strong it is. confidence assessment:
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12:19:06 ** Voltage is the work done per unit of charge in moving charge from one point to another. **
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RESPONSE --> ok self critique assessment:
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y|iܶ assignment #026 026. Query 27 Physics II 10-26-2008
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13:39:13 Note that the solutions given here use k notation rather than epsilon0 notation. The k notation is more closely related to the geometry of Gauss' Law and is consistent with the notation used in the Introductory Problem Sets. However you can easily switch between the two notations, since k = 1 / (4 pi epsilon0), or alternatively epsilon0 = 1 / (4 pi k).
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RESPONSE --> Okay I will stick with constant k
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13:40:22 Introductory Problem Set 2
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RESPONSE --> I don't know what you want me to say about it. Here is a formula for it. 1.6 * 10 ^ -19 Coulombs / electron * (nElectrons * `dL / L)
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13:51:45 Based on what you learned from Introductory Problem Set 2, how does the current in a wire of a given material, for a given voltage, depend on its length and cross-sectional area?
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RESPONSE --> A current is found by diving voltage over length times cross sectional area
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13:52:37 How can the current in a wire be determined from the drift velocity of its charge carriers and the number of charge carriers per unit length?
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RESPONSE --> I don't know what drift velocity is so I don't see how that would differ from the original equation.
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13:55:09 Will a wire of given length and material have greater or lesser electrical resistance if its cross-sectional area is greater, and why?
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RESPONSE --> Where is this in the textbook?
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13:56:15 Will a wire of given material and cross-sectional area have greater or lesser electrical resistance if its length is greater, and why?
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RESPONSE --> It will have less resistance is the length is greater
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13:56:17 Query Principles and General Physics 16.24: Force on proton is 3.75 * 10^-14 N toward south. What is magnitude and direction of the field?
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13:56:19 The direction of the electric field is the same as that as the force on a positive charge. The proton has a positive charge, so the direction of the field is to the south. The magnitude of the field is equal to the magnitude of the force divided by the charge. The charge on a proton is 1.6 * 10^-19 Coulombs. So the magnitude of the field is E = F / q = 3.75 * 10^-14 N / (1.6 * 10^-19 C) = 2.36* 10^5 N / C.
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13:56:21 Query gen phy problem 16.32. field 745 N/C midway between two equal and opposite point charges separated by 16 cm. What is the magnitude of each charge?
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RESPONSE -->
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13:56:23 ** If the magnitude of the charge is q then the field contribution of each charge is k q / r^2, with r = 8 cm = .08 meters. Since both charges contribute equally to the field, with the fields produced by both charges being in the same direction (on any test charge at the midpoint one force is of repulsion and the other of attraction, and the charges are on opposite sides of the midpoint), the field of either charge has magnitude 1/2 (745 N/C) = 373 N/C. Thus E = 373 N/C and E = k q / r^2. We know k, E and r so we solve for q to obtain q = E * r^2 / k = 373 N/C * (.08 m)^2 / (9 * 10^9 N m^2 / C^2) = 373 N/C * .0064 m^2 / (9 * 10^9 N m^2 / C^2) = 2.6 * 10^-10 C, approx. **
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13:56:25 If the charges are represented by Q and -Q, what is the electric field at the midpoint?
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13:56:27 ** this calls for a symbolic expression in terms of the symbol Q. The field would be 2 k Q / r^2, where r=.08 meters and the factor 2 is because there are two charges of magnitude Q both at the same distance from the point. **
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13:56:30 Query Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C charge.
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13:56:32 A positive charge at the given point will be repelled by the given positive charge, so will experience a force which is directly upward. The field has magnitude E = (k q Q / r^2) / Q, where q is the given charge and Q an arbitrary test charge introduced at the point in question. Since (k q Q / r^2) / Q = k q / r^2, we obtain E = 9 * 10^9 N m^2 / C^2 * 33.0 * 10^-6 C / (.200 m)^2 = 7.43 * 10^6 N / C.
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14:04:47 query univ 22.30 (10th edition 23.30). Cube with faces S1 in xz plane, S2 top, S3 right side, S4 bottom, S5 front, S6 back on yz plane. E = -5 N/(C m) x i + 3 N/(C m) z k. What is the flux through each face of the cube, and what is the total charge enclosed by the cube?
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RESPONSE --> Given E = (-5.00N/C)*mx component i + (3.00N/C)*m)z component k L = 0.300m component ns = +k which would be flux E*componentns A = (3.00(N/C) * m) (0.300m)^2z - (0.27N/C)m)z = (0.27(N/C)m(0.300m) = 0.081(N/C)m^2 ns = +component j flux = E * component ns A = 0 ns = -component k flux = E*component ns A = -(0.27(N/C)*m)z = 0 (z=0) ns = +component i flux = E*component ns A = (-5.00(N/C)*m)^2x = -(0.45(N/C)*m)x = - (0.45(N/C) * m().300m) = -().135 N/C*m^2 ns = -component i flux = E*component ns A = +(0.45(N/c) * m)x = 0 (x = 0) Total flux flux = flux + flux = (0.081 - 0.135)(N/C) * m^2 = -0.054Nm^2/C q = E0fluv = -4.78 x 10^-13C
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14:05:13 **** Advance correction of a common misconception: Flux is not a vector quantity so your flux values will not be multiples of the i, j and k vectors. The vectors normal to S1, S2, ..., S6 are respectively -j, k, j, -k, i and -i. For any face the flux is the dot product of the field with the normal vector, multiplied by the area. The area of each face is (.3 m)^2 = .09 m^2 So we have: For S1 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-j) * .09 m^2 = 0. For S2 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( k) * .09 m^2 = 3 z N / (C m) * .09 m^2. For S3 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( j) * .09 m^2 = 0. For S4 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-k) * .09 m^2 = -3 z N / (C m) * .09 m^2. For S5 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( i) * .09 m^2 = -5 x N / (C m) * .09 m^2. For S6 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-i) * .09 m^2 = 5 x N / (C m) * .09 m^2. On S2 and S4 we have z = .3 m and z = 0 m, respectively, giving us flux .027 N m^2 / C on S2 and flux 0 on S4. On S5 and S6 we have x = .3 m and x = 0 m, respectively, giving us flux -.045 N m^2 / C on S5 and flux 0 on S6. The total flux is therefore .027 N m^2 / C - .045 N m^2 / C = -.018 N m^2 / C. Since the total flux is 4 pi k Q, where Q is the charge enclosed by the surface, we have 4 pi k Q = -.018 N m^2 / C and Q = -.018 N m^2 / C / (4 pi k) = -.018 N m^2 / C / (4 pi * 9 * 10^9 N m^2 / C^2) = -1.6 * 10^-13 C, approx. **
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RESPONSE --> I'm not sure what I did.
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14:07:26 query univ 22.37 (23.27 10th edition) Spherical conducting shell inner radius a outer b, concentric with larger conducting shell inner radius c outer d. Total charges +2q, +4q. Give your solution.
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RESPONSE --> .Part A) r < R, E = 0 because there is no charge in the shell Part B) R < r < 2R, E = 1/4piE0 * Q/r since chargeis enclise is Qr > 2R E = 1/4piE0 * 2Q/r, since charge is enclosed is 2Q
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14:07:44 ** The electric field inside either shell must be zero, so the charge enclosed by any sphere concentric with the shells and lying within either shell must be zero, and the field is zero for a < r < b and for c < r < d. Thus the total charge on the inner surface of the innermost shell is zero, since this shell encloses no charge. The entire charge 2q of the innermost shell in concentrated on its outer surface. For any r such that b < r < c the charge enclosed by the corresponding sphere is the 2 q of the innermost shell, so that the electric field is 4 pi k * 2q / r^2 = 8 pi k q / r^2. Considering a sphere which encloses the inner but not the outer surface of the second shell we see that this sphere must contain the charge 2q of the innermost shell. Since this sphere is within the conducting material the electric field on this sphere is zero and the net flux thru this sphere is zero. Thus the total charge enclosed by this sphere is zero. Since the charge enclosed by the sphere includes the 2q of the innermost shell, the sphere must also enclose a charge -2 q, which by symmetry must be evenly distributed on the inner surface of the second shell. Any sphere which encloses both shells must enclose the total charge of both shells, which is 6 q. Since we have 2q on the innermost shell and -2q on the inner surface of the second shell the charge on the outer surface of this shell must be 6 q. For any r such that d < r the charge enclosed by the corresponding sphere is the 6 q of the two shells, so that the electric field is 4 pi k * 6q / r^2 = 24 pi k q / r^2. **
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RESPONSE --> This was just a confusing problem.
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14:10:53 query univ 23.46 (23.34 10th edition). Long conducting tube inner radius a, outer b. Lin chg density `alpha. Line of charge, same density along axis.
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RESPONSE --> Ex = (2alambda)/((4piE0zxa(sqroot(1+x^2.a^2) Ex = 1/2piE0 * lambda / x(sqroot(1+x^2/a^2)) Part B) The potential was found at y and z equal to zero, so there is no dependence on them./
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14:10:59 **The Gaussian surfaces appropriate to this configuration are cylinders of length L which are coaxial with the line charge. The symmetries of the situation dictate that the electric field is everywhere radial and hence that the field passes through the curved surface of each cylinder at right angle to that surface. The surface area of the curved portion of any such surface is 2 pi r L, where r is the radius of the cylinder. For r < a the charge enclosed by the Gaussian surface is L * alpha so that the flux is charge enclosed = 4 pi k L * alpha and the electric field is electric field = flux / area = 4 pi k L * alpha / (2 pi r L ) = 2 k alpha / r. For a < r < b, a Gaussian surface of radius r lies within the conductor so the field is zero (recall that if the field wasn't zero, the free charges inside the conductor would move and we wouldn't be in a steady state). So the net charge enclosed by this surface is zero. Since the line charge enclosed by the surface is L * alpha, the inner surface of the conductor must therefore contain the equal and opposite charge -L * alpha, so that the inner surface carries charge density -alpha. For b < r the Gaussian surface encloses both the line charge and the charge of the cylindrical shell, each of which has charge density alpha, so the charge enclosed is 2 L * alpha and the electric field is radial with magnitude 4 pi k * 2 L * alpha / (2 pi r L ) = 4 k alpha / r. Since the enclosed charge that of the line charge (L * alpha) as well as the inner surface of the shell (L * (-alpha) ), which the entire system carries charge L * alpha, we have line charge + charge on inner cylinder + charge on outer cylinder = alpha * L, we have alpha * L - alpha * L + charge on outer cylinder = alpha * L, so charge on outer cylinder = 2 alpha * L, so the outer surface of the shell has charge density 2 alpha. **
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RESPONSE --> ok
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ԆFжhpŕȍv assignment #027 027. Physics II 10-26-2008
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15:00:09 Query Principles and General Physics 17.4: work by field on proton from potential +135 V to potential -55 V.
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RESPONSE -->
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15:00:11 The change in potential is final potential - initial potential = -55 V - (135 V) = -190 V, so the change in the potential energy of the proton is -190 V * 1.6 * 10^-19 C = -190 J / C * 1.6 * 10^-19 C = -3.0 * 10^-17 J. In the absence of dissipative forces this is equal and opposite to the change in the KE of the proton; i.e., the proton would gain 3.09 * 10^-17 J of kinetic energy. Change in potential energy is equal and opposite to the work done by the field on the charge, so the field does 3.0 * 10^-17 J of work on the charge. Since the charge of the proton is equal in magnitude to that of an electron, he work in electron volts would be 180 volts * charge of 1 electron= 180 eV.
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15:00:14 Query Principles and General Physics 17.8: Potential difference required to give He nucleus 65.0 keV of KE.
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15:00:16 65.0 keV is 65.0 * 10^3 eV, or 6.50 * 10^4 eV, of energy. The charge on a He nucleus is +2 e, where e is the charge on an electron. So assuming no dissipative forces, for every volt of potential difference, the He nuclues would gain 2 eV of kinetic energy. To gain 6.50 * 10^4 eV of energy the voltage difference would therefore be half of 6.50 * 10^4 voles, or 3.35 * 10^4 volts.
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15:00:18 Query gen phy text problem 17.18 potential 2.5 * 10^-15 m from proton; PE of two protons at this separation in a nucleus. What is the electrostatic potential at a distance of 2.5 * 10^-15 meters?
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15:00:20 STUDENT SOLUTION: For a part, to determine the electric potential a distance fo 2.5810^-15m away from a proton, I simply used the equation V = k q / r for electric potential for point charge: q = 1.60*10^-19C=charge on proton V = kq/r = 9.0*10^9N*m^2/C^2(1.60*10^-19C) / (2.5*10^-15m) = 5.8*10^5V. Part B was the more difficult portion of the problem. You have to consider a system that consists of two protons 2.5*10^-5m apart. The work done against the electric field to assemble these charges is W = qV. The potential energy is equal to the work done against the field. PE=(1.60*10^-19C)(5.8*10^5V) = 9.2*10^-14 J.
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15:15:47 query univ phy 23.58 (24.58 10th edition). Geiger counter: long central wire 145 microns radius, hollow cylinder radius 1.8 cm. What potential difference between the wire in the cylinder will produce an electric field of 20,000 volts/m at 1.2 cm from the wire?
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RESPONSE --> We need the potential difference V Vab = Elin(b/a) * r Vab = (2.00 x 10^-4N/C)(ln(0.018m/145 x 10^-6m)) * 0.012m = 1157 V
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15:16:07 ** The voltage V_ab is obtained by integrating the electric field from the radius of the central wire to the outer radius. From this we determine that E = Vab / ln(b/a) * 1/r, where a is the inner radius and b the outer radius. If E = 20,000 V/m at r = 1.2 cm then Vab = E * r * ln(b/a) = 20,000 V/m * ln(1.8 cm / .0145 cm) * .012 m = 1157 V. **
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RESPONSE --> awesome.
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15:18:56 Query univ 23.78 (24.72 10th edition). Rain drop radius .65 mm charge -1.2 pC. What is the potential at the surface of the rain drop?
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RESPONSE --> First we need to find voltage(or potential difference V) V = kQ / r V = k (-1.20 x 10^-12 C) / 6.50 x 10^-4m V = -16.6v Part B) The volume doubles, so the radius increases by the cube root of 2. Rnew = 8.19 x 10^-4m Qnew = 2Q = -2.40 x 10^-12 C So the new potential V is: Vnew = kQnew / Rnew V = k(-2.40 x 10^-12 C) / 8.19 x 10^-4 m V = -26.4V
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15:19:42 STUDENT RESPONSE FOLLOWED BY SOLUTION: The problem said that V was 0 at d = inifinity, which I understnad to mean that as we approach the raindrop from infinity, the potential differencegrows from 0, to some amount at the surface of the raindrop. Because water molecules are more positive on one side that the other, they tend to align in a certain direction. Since positive charges tend to drift toward negative charge, I would think that the raindrop, with its overall negative charge, has molecules arranged so that their more positive sides are pointing toward the center and negative sides will be alighed along the surface of the raindrop. Probably all wrong. I tried several differnet integrand configuraitons but never found one that gave me an answer in volts. SOLUTION: You will have charge Q = -1.2 * 10^-12 C on the surface of a sphere of radius .00065 m. The field is therefore E = k Q / r^2 = 9 * 10^9 N m^2 / C^2 * (-1.2 * 10^-12 C) / r^2 = -1.08 * 10^-2 N m^2 / C / r^2. Integrating the field from infinity to .00065 m we get (-1.08 * 10^-2 N m^2 / C) / (.00065 m) = -16.6 N m / C = -16.6 V. If two such drops merge they form a sphere with twice the volume and hence 2^(1/3) times the radius, and twice the charge. The surface potential is proportional to charge and inversely proportional to volume. So the surface potential will be 2 / 2^(1/3) = 2^(2/3) times as great as before. The surface potential is therefore 16.6 V * 2^(2/3) = -26.4 volts, approx.. **
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RESPONSE --> I think I am getting it.
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`ܽ{̳zT assignment #028 028. `Query 28 Physics II 10-26-2008
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15:21:26 Query introductory problems set 54 #'s 1-7. Explain how to obtain the magnetic field due to a given current through a small current segment, and how the position of a point with respect to the segment affects the result.
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RESPONSE --> We use the formula with the electric field strength E. E = kq/r^2. With this we can find the strength of the electric field.
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15:22:58 ** IL is the source. The law is basically an inverse square law and the angle theta between IL and the vector r from the source to the point also has an effect so that the field is B = k ' I L / r^2 * sin(theta). **
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RESPONSE --> Is B the strength of the magnetic field and E is the strength of the electric field?
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15:23:01 Query principles and general college physics problem 17.34: How much charge flows from each terminal of 7.00 microF capacitor when connected to 12.0 volt battery?
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15:23:03 Capacitance is stored charge per unit of voltage: C = Q / V. Thus the stored charge is Q = C * V, and the battery will have the effect of transferring charge of magnitude Q = C * V = 7.00 microF * 12.0 volts = 7.00 C / volt * 12.0 volts = 84.0 C of charge. This would be accomplished the the flow of 84.0 C of positive charge from the positive terminal, or a flow of -84.0 C of charge from the negative terminal. Conventional batteries in conventional circuits transfer negative charges.
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15:26:33 Explain how to obtain the magnetic field due to a circular loop at the center of the loop.
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RESPONSE --> I can rearrange the following equation but when we have a circle: R = (mv)/ qB or B = (mv) / q Where B is the strength of the magnetic field, m is the mass of the particle, v is the velocity, and q is the charge.
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15:28:15 ** For current running in a circular loop: Each small increment `dL of the loop is a source I `dL. The vector from `dL to the center of the loop has magnitude r, where r is the radius of the loop, and is perpendicular to the loop so the contribution of increment * `dL to the field is k ' I `dL / r^2 sin(90 deg) = I `dL / r^2, where r is the radius of the loop. The field is either upward or downward by the right-hand rule, depending on whether the current runs counterclockwise or clockwise, respectively. The field has this direction regardless of where the increment is located. The sum of the fields from all the increments therefore has magnitude B = sum(k ' I `dL / r^2), where the summation occurs around the entire loop. I and r are constants so the sum is B = k ' I / r^2 sum(`dL). The sum of all the length increments around the loop is the circumference 2 pi r of the loop so we have B = 2 pi r k ' I / r^2 = 2 pi k ' I / r. **
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RESPONSE --> okay we are talking about completely different things. I am getting some concepts right with electricity but I am getting some things way off target. Sometimes I do not know what we are talking about since we are talking about so much at once.
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15:28:41 Query magnetic fields produced by electric currents. What evidence do we have that electric currents produce magnetic fields?
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RESPONSE --> I am not sure. I know they are perpendicular to each other.
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15:29:57 STUDENT RESPONSE: We do have evidence that electric currents produce magnetic fields. This is observed in engineering when laying current carrying wires next to each other. The current carrying wires produce magnetic fields that may affect other wires or possibly metal objects that are near them. This is evident in the video experiment. When Dave placed the metal ball near the coil of wires and turned the generator to produce current in the wires the ball moved toward the coil. This means that there was an attraction toward the coil which in this case was a magnetic field. INSTRUCTOR COMMENT: Good observations. A very specific observation that should be included is that a compass placed over a conducting strip or wire initially oriented in the North-South direction will be deflected toward the East-West direction. **
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RESPONSE --> When I talked about them being perpendicular, I did not think about the magnets.
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15:30:44 How is the direction of an electric current related to the direction of the magnetic field that results?
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RESPONSE --> they are perpendicular to each other. In the last box we talked about relations between North South magnet and the East West direction.
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15:31:28 ** GOOD STUDENT RESPONSE: The direction of the magnetic field relative to the direction of the electric current can be described using the right hand rule. This means simply using your right hand as a model you hold it so that your thumb is extended and your four fingers are flexed as if you were holding a cylinder. In this model, your thumb represents the direction of the electric current in the wire and the flexed fingers represent the direction of the magnetic field due to the current. In the case of the experiment the wire was in a coil so the magnetic field goes through the hole in the middle in one direction. **
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15:31:32 Query problem 17.35 What would be the area of a .20 F capacitor if plates are separated by 2.2 mm of air?
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15:31:35 ** If a parallel-plate capacitor holds charge Q and the plates are separated by air, which has dielectric constant very close to 1, then the electric field between the plates is E = 4 pi k sigma = 4 pi k Q / A, obtained by applying Gauss' Law to each plate. The voltage between the plates is therefore V = E * d, where d is the separation. The capacitance is C = Q / V = Q / (4 pi k Q / A * d) = A / (4 pi k d). Solving this formula C = A / (4 pi k d) for A for the area we get A = 4 pi k d * C If capacitance is C = .20 F and the plates are separated by 2.2 mm = .0022 m we therefore have A = 4 pi k d * C = 4 pi * 9 * 10^9 N m^2 / C^2 * .20 C / V * .0022 m = 5 * 10^7 N m^2 / C^2 * C / ( J / C) * m = 5 * 10^7 N m^2 / (N m) * m = 5 * 10^7 m^2. **
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15:31:38 Query problem 17.50 charge Q on capacitor; separation halved as dielectric with const K inserted; by what factor does the energy storage change? Compare the new electric field with the old.
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15:31:40 Note that the problem in the latest version of the text doubles rather than halves the separation. The solution for the halved separation, given here, should help you assess whether your solution was correct, and if not should help you construct a detailed and valid self-critique. ** For a capacitor we know the following: Electric field is independent of separation, as long as we don't have some huge separation. Voltage is work / unit charge to move from one plate to the other, which is force / unit charge * distance between plates, or electric field * distance. That is, V = E * d. Capacitance is Q / V, ration of charge to voltage. Energy stored is .5 Q^2 / C, which is just the work required to move charge Q across the plates with the 'average' voltage .5 Q / C (also obtained by integrating `dW = `dQ * V = `dq * q / C from q=0 to q = Q). The dielectric increases capacitance by reducing the electric field, which thereby reduces the voltage between plates. The electric field will be 1 / k times as great, meaning 1/k times the voltage at any given separation. C will increase by factor k due to dielectric, and will also increase by factor 2 due to halving of the distance. This is because the electric field is independent of the distance between plates, so halving the distance will halve the voltage between the plates. Since C = Q / V, this halving of the denominator will double C. Thus the capacitance increases by factor 2 k, which will decrease .5 Q^2 / C, the energy stored, by factor 2 k. **
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15:37:20 query univ 24.50 (25.36 10th edition). Parallel plates 16 cm square 4.7 mm apart connected to a 12 volt battery. What is the capacitance of this capacitor?
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RESPONSE --> Part A) Capacitance C = (E0*A)/ d = E0(0.16m)^2 / 4.7 x 10^-3m C = 4.8 x 10^-11 F Part B) Q = CV = (4.8 x 10^-11 F)(12 V) = 0.58 x 10^-9 C Part C) E = V/d E = 12V / 4.7 x 10^-3, E = 2553 V / m Part D) U = 1/2 CV^2 = 1/2 (4.8 x 10^-11F)(12V)^2 = 3.46 x 10^-9J Part E) If the battery is disconnected, the charge remains constant, and the plates are pulled apart to 0.0094m. I basically repeat parts A-D and here would be my answers using the exact same formulas. C = 2.41 x 10^-11 F Q = 0.58 x 10^-9 C E = 2553 V/m U = 6.91 x 10^-9 J
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15:37:42 ** Fundamental principles include the fact that the electric field is very neary constant between parallel plates, the voltage is equal to field * separation, electric field from a single plate is 2 pi k sigma, the work required to displace a charge is equal to charge * ave voltage, and capacitance is charge / voltage. Using these principles we reason out the problem as follows: If the 4.7 mm separation experiences a 12 V potential difference then the electric field is E = 12 V / (4.7 mm) = 12 V / (.0047 m) = 2550 V / m, approx. Since the electric field of a plane charge distribution with density sigma is 2 pi k sigma, and since the electric field is created by two plates with equal opposite charge density, the field of the capacitor is 4 pi k sigma. So we have 4 pi k sigma = 2250 V / m and sigma = 2250 V / m / (4 pi k) = 2250 V / m / (4 pi * 9 * 10^9 N m^2 / C^2) = 2.25 * 10^-8 C / m^2. The area of the plate is .0256 m^2 so the charge on a plate is .0256 m^2 * 2.25 * 10^-8 C / m^2 = 5.76 * 10^-10 C. The capacitance is C = Q / V = 5.67 * 10^-10 C / (12 V) = 4.7 * 10^-11 C / V = 4.7 * 10^-11 Farads. The energy stored in the capacitor is equal to the work required to move this charge from one plate to another, starting with an initially uncharged capacitor. The work to move a charge Q across an average potential difference Vave is Vave * Q. Since the voltage across the capacitor increases linearly with charge the average voltage is half the final voltage, so we have vAve = V / 2, with V = 12 V. So the energy is energy = vAve * Q = 12 V / 2 * (5.76 * 10^-10 C) = 3.4 * 10^-9 V / m * C. Since the unit V / m * C is the same as J / C * C = J, we see that the energy is 3.4 * 10^-9 J. Pulling the plates twice as far apart while maintaining the same voltage would cut the electric field in half (the voltage is the same but charge has to move twice as far). This would imply half the charge density, half the charge and therefore half the capacitance. Since we are moving only half the charge through the same average potential difference we use only 1/2 the energy. Note that the work to move charge `dq across the capacitor when the charge on the capacitor is `dq * V = `dq * (q / C), so to obtain the work required to charge the capacitor we integrate q / C with respect to q from q = 0 to q = Q, where Q is the final charge. The antiderivative is q^2 / ( 2 C ) so the definite integral is Q^2 / ( 2 C). This is the same result obtained using average voltage and charge, which yields V / 2 * Q = (Q / C) / 2 * Q = Q^2 / (2 C) Integration is necessary for cylindrical and spherical capacitors and other capacitors which are not in a parallel-plate configuration. **
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RESPONSE --> Again I really have no idea if what I am doing is right or not.
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15:40:10 query univ 24.51 (25.37 10th edition). Parallel plates 16 cm square 4.7 mm apart connected to a 12 volt battery. If the battery remains connected and plates are pulled to separation 9.4 mm then what are the capacitance, charge of each plate, electric field, energy stored?
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RESPONSE --> This time the voltage remains constant, and the plates are pulled apart just like at the end of the last problem. So using the same formulas we would find that: C = 2.4 x 10^-11 F Q = 2.9 x 10^-10 C E = V/d E= 12 V / 0.0094 E = 1.3 x10^3 V/m U = CV^2 / 2 U = (2.4 x 10^-11 F)(12 V)^2 / 2 U = 1.73 x 10^-9 J
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15:40:46 The potential difference between the plates is originally 12 volts. 12 volts over a 4.7 mm separation implies electric field = potential gradient = 12 V / (.0047 m) = 2500 J / m = 2500 N / C, approx.. The electric field is E = 4 pi k * sigma = 4 pi k * Q / A so we have Q = E * A / ( 4 pi k) = 2500 N / C * (.16 m)^2 / (4 * pi * 9 * 10^9 N m^2 / C^2) = 5.7 * 10^-7 N / C * m^2 / (N m^2 / C^2) = 5.7 * 10^-10 C, approx.. The energy stored is E = 1/2 Q V = 1/2 * 5.6 * 10^-10 C * 12 J / C = 3.36 * 10^-9 J. If the battery remains connected then if the plate separation is doubled the voltage will remain the same, while the potential gradient and hence the field will be halved. This will halve the charge on the plates, giving us half the capacitance. So we end up with a charge of about 2.8 * 10^-10 C, and a field of about 1250 N / C. The energy stored will also be halved, since V remains the same but Q is halved.
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RESPONSE --> I feel like sometimes I am getting this stuff and other times I'm just completely wrong.
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15:49:18 query univ 24.68 (25.52 10th edition). Solid conducting sphere radius R charge Q. What is the electric-field energy density at distance r < R from the center of the sphere? What is the electric-field energy density at distance r > R from the center of the sphere?
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RESPONSE --> Part A) r < R u = 1/2 E0 * E^2 = 0 Part B) r > R u = 1/2E0 * E^2 = 1/2E0 ((Q/4piE0r^2)^2 = Q^2 / 32pi^2 E0 r^4 Part C) We need to integrate uDV. I am not sure how to keep going with this problem Part D) Since i can;t solve part C. I can't solve D. Part E) I know I need to use this: U = Q/2C
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15:49:59 ** The idea is that we have to integrate the energy density over all space. We'll do this by finding the total energy in a thin spherical shell of radius r and thickness `dr, using this result to obtain an expression we integrate from R to infinity, noting that the field of the conducting sphere is zero for r < R. Then we can integrate to find the work required to assemble the charge on the surface of the sphere and we'll find that the two results are equal. Energy density, defined by dividing the energy .5 C V^2 required to charge a parallel-plate capacitor by the volume occupied by its electric field, is Energy density = .5 C V^2 / (volume) = .5 C V^2 / (d * A), where d is the separation of the plates and A the area of the plates. Since C = epsilon0 A / d and V = E * d this gives us .5 epsilon0 A / d * (E * d)^2 / (d * A) = .5 epsilon0 E^2 so that Energy density = .5 epsilon0 E^2, or in terms of k Energy density = 1 / (8 pi k) E^2, Since your text uses epsilon0 I'll do the same on this problem, where the epsilon0 notation makes a good deal of sense: For the charged sphere we have for r > R E = Q / (4 pi epsilon0 r^2), and therefore energy density = .5 epsilon0 E^2 = .5 epsilon0 Q^2 / (16 pi^2 epsilon0^2 r^4) = Q^2 / (32 pi^2 epsilon0 r^4). The energy density between r and r + `dr is nearly constant if `dr is small, with energy density approximately Q^2 / (32 pi^2 epsilon0 r^4). The volume of space between r and r + `dr is approximately A * `dr = 4 pi r^2 `dr. The expression for the energy lying between distance r and r + `dr is therefore approximately energy density * volume = Q^2 / (32 pi^2 epsilon0 r^4) * 4 pi r^2 `dr = Q^2 / (8 pi epsilon0 r^2) `dr. This leads to a Riemann sum over radius r; as we let `dr approach zero we approach an integral with integrand Q^2 / (8 pi epsilon0 r^2), integrated with respect to r. To get the energy between two radii we therefore integrate this expression between those two radii. If we integrate this expression from r = R to infinity we get the total energy of the field of the charged sphere. This integral gives us Q^2 / (8 pi epsilon0 R), which is the same as k Q^2 / (2 R). The work required to bring a charge `dq from infinity to a sphere containing charge q is k q / R `dq, leading to the integral of k q / R with respect to q. If we integrate from q = 0 to q = Q we get the total work required to charge the sphere. Our antiderivative is k (q^2 / 2) / r. If we evaluate this antiderivative at lower limit 0 and upper limit Q we get k Q^2 / (2 R). Since k = 1 / (4 pi epsilon0) the work is k Q^2 / (2 R) = k Q^2 / (8 pi epsilon0 R). So the energy in the field is equal to the work required to assemble the charge distribution. **
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RESPONSE --> I know we had to integrate and I can integrate uDV. But I had a problem trying to setup the limits of integration.
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Mͫǩfz assignment #029 029. `Query 29 Physics II 10-26-2008
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15:52:15 Query introductory problem set 54 #'s 8-13 Explain how to determine the magnetic flux of a uniform magnetic field through a plane loop of wire, and explain how the direction of the field and the direction of a line perpendicular to the plane of the region affect the result.
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RESPONSE --> To find direction we need to use the right hand rule. Wrap your fingers around the wire. Your thumb will be the direction of the flux. Magnetic flux is found by taking the product of the magnetic strength B and cross sectional area A.
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15:52:55 To do this we need to simply find the area of the plane loop of wire. If we are given the radius we can find the area using Pi * r ^2 Then we multiply the area of the loop (In square meters ) by the strength of the field (in tesla). This will give us the strength of the flux if the plane of the loop is perpendicular to the field. If the perpendicular to the loop is at some nonzero angle with the field, then we multiply the previous result by the cosine of the angle.
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RESPONSE --> Area of a circle is pi * r^2. (I did not specify that)
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15:53:53 Explain how to determine the average rate of change of magnetic flux due to a uniform magnetic field through a plane loop of wire, as the loop is rotated in a given time interval from an orientation perpendicular to the magnetic field to an orientation parallel to the magnetic field.
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RESPONSE --> For this one can we find the flux seperately and take the difference of them?
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15:57:34 ** EXPLANATION BY STUDENT: The first thing that we need to do is again use Pi * r ^ 2 to find the area of the loop. Then we multiply the area of the loop (m^2) by the strength of the field (testla) to find the flux when the loop is perpendicular to the field. Then we do the same thing for when the loop is parallel to the field, and since the cos of zero degrees is zero, the flux when the loop is parallel to the field is zero. This makes sense because at this orientation the loop will pick up none of the magnetic field. So now we have Flux 1 and Flux 2 being when the loop is perpendicular and parallel, respectively. So if we subtract Flux 2 from flux 1 and divide this value by the given time in seconds, we will have the average rate of change of magnetic flux. If we use MKS units this value will be in Tesla m^2 / sec = volts. **
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RESPONSE --> I see. I need to take the difference and then divide by time t
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15:59:03 Explain how alternating current is produced by rotating a coil of wire with respect to a uniform magnetic field.
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RESPONSE --> Current is not usually constant (unless it is in a direct series circuit) So it will vary in a coil of wire.
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16:04:08 ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: Y ou rotate a coil of wire end over end inside a uniform magnetic field. When the coil is parallel to the magnetic field, then there is no magnetic flux, and the current will be zero. But then when the coil is perpendicular to the field or at 90 degrees to the field then the flux will be strongest and the current will be moving in one direction. Then when the coil is parallel again at 180 degrees then the flux and the current will be zero. Then when the coil is perpendicular again at 270 degrees, then the flux will be at its strongest again but it will be in the opposite direction as when the coil was at 90 degrees. So therefore at 90 degrees the current will be moving in one direction and at 270 degrees the current will be moving with the same magnitude but in the opposite direction. COMMENT: Good. The changing magnetic flux produces voltage, which in turn produces current. **
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RESPONSE --> Thank you that makes a little more sense to me. Change in magnetic flux produces voltage, which also produces current.
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16:04:11 Query Principles and General College Physics 18.04. 120V toaster with 4.2 amp current. What is the resistance?
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16:04:13 current = voltage / resistance (Ohm's Law). The common sense of this is that for a given voltage, less resistance implies greater current while for given resistance, greater voltage implies greater current. More specifically, current is directly proportional to voltage and inversely proportional to resistance. In symbols this relationship is expressed as I = V / R. In this case we know the current and the voltage and wish to find the resistance. Simple algebra gives us R = V / I. Substituting our known current and voltage we obtain R = 120 volts / 4.2 amps = 29 ohms, approximately.
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16:04:15 Query Principles and General College Physics 18.28. Max instantaneous voltage to a 2.7 kOhm resistor rated at 1/4 watt.
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16:05:02 Voltage is energy per unit of charge, measured in Joules / Coulomb. Current is charge / unit of time, measured in amps or Coulombs / second. Power is energy / unit of time measured in Joules / second. The three are related in a way that is obvious from the meanings of the terms. If we multiply Joules / Coulomb by Coulombs / second we get Joules / second, so voltage * current = power. In symbols this is power = V * I. Ohm's Law tells us that current = voltage / resistance.In symbols this is I = V / R. So our power relationship power = V * I can be written power = V * V / R = V^2 / R. Using this relationship we find that V = sqrt(power * R), so in this case the maximum voltage (which will produce the 1/4 watt maximum power) will be V = sqrt(1/4 watt * 2.7 * 10^3 ohms) = 26 volts.
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16:05:05 Query general college physics problem 18.39; compare power loss if 520 kW delivered at 50kV as opposed to 12 kV thru 3 ohm resistance.** The current will not be the same at both voltages. It is important to understand that power (J / s) is the product of current (C / s) and voltage (J / C). So the current at 50 kV kW will be less than 1/4 the current at 12 kV. To deliver 520 kW = 520,000 J / s at 50 kV = 50,000 J / C requires current I = 520,000 J/s / (50,000 J/C) = 10.4 amps. This demonstrates the meaning of the formula P = I V. To deliver 520 kW = 520,000 J / s at 12 kV = 12,000 J / C requires current I = 520,000 J/s / (12,000 J/C) = 43.3 amps. The voltage drops through the 3 Ohm resistance will be calculated as the product of the current and the resistance, V = I * R: The 10.4 amp current will result in a voltage drop of 10.4 amp * 3 ohms = 31.2 volts. The 43.3 amp current will result in a voltage drop of 40.3 amp * 3 ohms = 130 volts. The power loss through the transmission wire is the product of the voltage ( J / C ) and the current (J / S) so we obtain power losses as follows: At 520 kV the power loss is 31.2 J / C * 10.4 C / s = 325 watts, approx. At 12 kV the power loss is 130 J / C * 43.3 C / s = 6500 watts, approx. Note that the power loss in the transmission wire is not equal to the power delivered by the circuit, which is lost through a number of parallel connections to individual homes, businesses, etc.. The entire analysis can be done by simple formulas but without completely understanding the meaning of voltage, current, resistance, power and their relationships it is very easy to get the wrong quantities in the wrong places, and especially to confuse the power delivered with the power loss. The analysis boils down to this: I = P / V, where P is the power delivered. Ploss = I^2 R, where R is the resistance of the circuit and Ploss is the power loss of the circuit. So Ploss = I^2 * R = (P/V)^2 * R = P^2 * R / V^2. This shows that power loss across a fixed resistance is inversely proportional to square of the voltage. So that the final voltage, which is less than 1/4 the original voltage, implies more than 16 times the power loss. A quicker solution through proportionalities: For any given resistance power loss is proportional to the square of the current. For given power delivery current is inversely proportional to voltage. So power loss is proportional to the inverse square of the voltage. In this case the voltage ratio is 50 kV / (12 kV) = 4.17 approx., so the ratio of power losses is about 1 / 4.17^2 = 1 / 16.5 = .06. Note that this is the same approximate ratio you would get if you divided your 324.5 watts by 5624.7 watts. **
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16:05:07 Query univ 25.62 (26.50 10th edition) rectangular block d x 2d x 3d, potential difference V. To which faces should the voltage be applied to attain maximum current density and what is the density?
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16:05:10 ** First note that the current I is different for diferent faces. The resistance of the block is proportional to the distance between faces and inversely proportional to the area, so current is proportional to the area and inversely proportional to the distance between faces. Current density is proportional to current and inversely proportional to the area of the face, so current density is proportional to area and inversely proportional to the distance between faces and to area, leaving current inversely proportional to distance between faces. For the faces measuring d x 2d we have resistance R = rho * L / A = rho * (3d) / (2 d^2) = 3 / 2 rho / d so current is I = V / R = V / (3/2 rho / d) = 2d V / (3 rho). Current density is I / A = (2 d V / (3 rho) ) / (2 d^2) = V / (3 rho d) = 1/3 V / (rho d). For the faces measuring d x 3d we have resistance R = rho * L / A = rho * (2d) / (3 d^2) = 2 / 3 rho / d so current is I = V / R = V / (2/3 rho / d) = 3 d V / (2 rho). Current density is I / A = (3 d V / (2 rho) ) / (3 d^2) = V / (2 rho d) = 1/2 V / (rho d). For the faces measuring 3d x 2d we have resistance R = rho * L / A = rho * (d) / (6 d^2) = 1 / 6 rho / d so current is I = V / R = V / (1/6 rho / d) = 6 d V / (rho). Current density is I / A = (6 d V / (rho) ) / (6 d^2) = V / (rho d) = V / (rho d). Max current density therefore occurs when the voltage is applied to the largest face. **
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dDzɶ۽ۇw{]ވ assignment #030 030. `Query 30 Physics II 10-26-2008
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16:11:36 Query introductory problem set 54 #'s 14-18. Explain whether, and if so how, the force on a charged particle due to the field between two capacitor plates is affected by its velocity.
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RESPONSE --> If a particle is inside a field for a shorter time(meaning it is traveling faster), then it has less time to be pulled to one side or the other. When a particle enters a magnetic field it will move in a semicircular path(it will not do this completely because it will generally get out of the field by then). By having a higher velocity, the magnetic field between the two capacitors has less time to apply a force on a charged particle.
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16:13:18 ** There is a force due to the electric field between the plates, but the effect of an electric field does not depend on velocity. The plates of a capacitor do not create a magnetic field. **
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RESPONSE --> I thought it would depend on velocity. Regardless there will be a force and an affect will take place whether the particle is moving 1 m/sec or 1 x 10^7m/sec. But for how long depends on the velocity(I am thinking of the equation 'ds = v* 'dt)
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16:14:05 Explain whether, and if so how, the force on a charged particle due to the magnetic field created by a wire coil is affected by its velocity.
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RESPONSE --> Apparently it is not. Velocity will only affect how long there will be a force on a charged particle. It will not affect whether there is one or not.
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16:14:47 ** A wire coil does create a magnetic field perpendicular to the plane of the coil. If the charged particle moves in a direction perpendicular to the coil then a force F = q v B is exerted by the field perpendicular to both the motion of the particle and the direction of the field. The precise direction is determined by the right-hand rule. **
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RESPONSE --> oh okay so velocity would be a factor since force is found by taking the product of charge, velocity, and the strength of the magnetic field.
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16:15:54 Explain how the net force changes with velocity as a charged particle passes through the field between two capacitor plates, moving perpendicular to the constant electric field, in the presence of a constant magnetic field oriented perpendicular to both the velocity of the particle and the field of the capacitor.
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RESPONSE --> Whenever a charged particle enters a magnetic field, it will change velocity because it will change direction. It will start to move in a semi-circular path until it is out of the field where it will move as it did before (unless some other force is going to be acting on it)
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16:16:39 ** At low enough velocities the magnetic force F = q v B is smaller in magnitude than the electrostatic force F = q E. At high enough velocities the magnetic force is greater in magnitude than the electrostatic force. At a certain specific velocity, which turns out to be v = E / B, the magnitudes of the two forces are equal. If the perpendicular magnetic and electric fields exert forces in opposite directions on the charged particle then when the magnitudes of the forces are equal the net force on the particle is zero and it passes through the region undeflected. **
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RESPONSE --> okay, So it depends on how high or low the velocity is(it depends on the magnitude of the velocity.)
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16:16:42 Query Principles and General Physics 20.2: Force on wire of length 160 meters carrying 150 amps at 65 degrees to Earth's magnetic field of 5.5 * 10^-5 T.
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16:16:44 The force on a current is I * L * B sin(theta) = 150 amps * 160 meters * 5.5 * 10^-5 T * sin(65 deg) = 1.20 amp * m * (N / (amp m) ) = 1.20 Newtons. Note that a Tesla, the unit of magnetic field, has units of Newtons / (amp meter), meaning that a 1 Tesla field acting perpendicular to a 1 amp current in a carrier of length 1 meters produces a force of 1 Newton. The question didn't ask, but be sure you know that the direction of the force is perpendicular to the directions of the current and of the field, as determined by the right-hand rule (fingers in direction of current, hand oriented to 'turn' fingers toward field, thumb in direction of force).
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16:16:46 Query Principles and General Physics 20.10. Force on electron at 8.75 * 10^5 m/s east in vertical upward magnetic field of .75 T.
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16:16:48 The magnitude of the force on a moving charge, exerted by a magnetic field perpendicular to the direction of motion, is q v B, where q is the charge, v the velocity and B the field. The force in this case is therefore F = q v B = 1.6 * 10^-19 C * 8.75 * 10^5 m/s * .75 T = 1.05 * 10^-13 C m/s * T = 1.05 * 10^-13 N. (units analysis: C m/s * T = C m/s * (N / (amp m) ) = C m/s * (kg m/s^2) / ((C/s) * m), with all units expressed as fundamental units. The C m/s in the numerator 'cancels' with the C m/s in the denominator, leaving kg m/s^2, or Newtons). The direction of the force is determined by the right-hand rule (q v X B) with the fingers in the direction of the vector q v, with the hand oriented to turn the fingers toward the direction of B. The charge q of the electron is negative, so q v will be in the direction opposite v, to the west. In order for the fingers to 'turn' qv toward B, the palm will therefore be facing upward, the fingers toward the west, so that the thumb will be pointing to the north. The force is therefore directed to the north.
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16:16:52 Query General Physics Problem (formerly 20.32, but omitted from new version). This problem is not assigned but you should solve it now: If an electron is considered to orbit a proton in a circular orbit of radius .529 * 10^-10 meters (the electron doesn't really move around the proton in a circle; the behavior of this system at the quantum level does not actually involve a circular orbit, but the result obtained from this assumption agrees with the results of quantum mechanics), the electron's motion constitutes a current along its path. What is the field produced at the location of the proton by the current that results from this 'orbit'? To obtain an answer you might want to first answer the two questions: 1. What is the velocity of the electron? 2. What therefore is the current produced by the electron? How did you calculate the magnetic field produced by this current?
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16:16:55 **If you know the orbital velocity of the electron and orbital radius then you can determine how long it takes to return to a given point in its orbit. So the charge of 1 electron 'circulates' around the orbit in that time interval. Current is charge flowing past a point / time interval. Setting centripetal force = Coulomb attraction for the orbital radius, which is .529 Angstroms = .529 * 10^-10 meters, we have m v^2 / r = k q1 q2 / r^2 so that v = sqrt(k q1 q2 / (m r) ). Evaluating for k, with q1 = q2 = fundamental charge and m = mass of the electron we obtain v = 2.19 * 10^6 m/s. The circumference of the orbit is `dt = 2 pi r so the time required to complete an orbit is `dt = 2 pi r / v, which we evaluate for the v obtained above. We find that `dt = 1.52 * 10^-16 second. Thus the current is I = `dq / `dt = q / `dt, where q is the charge of the electron. Simplifying we get I = .00105 amp, approx.. The magnetic field due to a .00105 amp current in a loop of radius .529 Angstroms is B = k ' * 2 pi r I / r^2 = 2 pi k ' I / r = 12.5 Tesla. **
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16:22:54 query univ 27.60 (28.46 10th edition). cyclotron 3.5 T field.
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RESPONSE --> K = 2.7MeV (2.7 x 10^6 eV)(1.6 x 10^-19J/eV) = 4.32 x 10^-13J v = sqroot(2K / m) = sqroot(2(4.32 x 10^-13J)/ 1.67 x 10^-27 kg) = 2.27 x 10^7m/sec R = mv / qB = ((1.67 x 10^-27 kg)(2.27 x 10^7 m/sec) / (1.6 x 10^-19 C)(3.5 T) R = 0.068 m omega = v / R omega = 2.27 x 10^7 m /sec / 0.068m omega = 3.34 x 10^8 rad/sec Part B) If the energy reaches the final value of 5.4 MeV, the velocity increase and so would the radius.
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16:23:05 What is the radius of orbit for a proton with kinetic energy 2.7 MeV?
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RESPONSE --> I found it already.
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16:23:33 ** We know that the centripetal force for an object moving in a circle is F = m v^2 / r. In a magnetic field perpendicular to the velocity this force is equal to the magnetic force F = q v B. So we have m v^2 / r = q v B so that r = m v / (q B). A proton with ke 2.7 MeV = 2.7 * 10^6 * (1.6 * 10^-19 J) = 3.2 * 10^-13 J has velocity such that v = sqrt(2 KE / m) = sqrt(2 * 3.2 * 10^-13 J / (1.67 * 10^-27 kg) ) = 2.3 * 10^7 m/s approx.. So we have r = m v / (q B) = 1.67 * 10^-27 kg * 2.3 * 10^7 m/s / (1.6 * 10^-19 C * 3.5 T) = .067 m approx. **
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RESPONSE --> It's very cool to combine the other equations in that way.
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16:23:46 What is the radius of orbit for a proton with kinetic energy 5.4 MeV?
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RESPONSE --> I think I already found this too?
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16:24:32 ** Doubling the KE of the proton increases velocity by factor sqrt(2) and therefore increases the radius of the orbit by the same factor. We end up with a radius of about .096 m. **
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RESPONSE --> Oh wait I did not put that down when I referred to the final valye of 5.4MeV. The velocity increases by sqroot of 2 and the radius increases to 0.096m. omega remains the same.
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16:27:40 query univ 28.73 (was 28.52) rail gun bar mass m with current I across rails, magnetic field B perpendicular to loop formed by bars and rails What is the expression for the magnitude of the force on the bar, and what is the direction of the force?
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RESPONSE --> We use ampere's law: B(2pi*r) = micro0 I
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16:27:57 ** The length of the bar is given as L. So the force is I L B, since the current and field are perpendicular. The acceleration of the bar is therefore a = I L B / m. If the distance required to achieve a given velocity is `ds and initial velocity is 0 then vf^2 = v0^2 + 2 a `ds gives us ds = (vf^2 - v0^2) / (2 a) = vf^2 / (2 a). If v stands for the desired final velocity this is written `ds = v^2 / (2 a). In terms of I, L, B and m we have `ds = v^2 / (2 I L B / m) = m v^2 / (2 I L B). Note that we would get the same expression using KE: since (neglecting dissipative losses) we have `dKE = `dW = F `ds we have `ds = `dKE / F = 1/2 m v^2 / (I L B). For the given quantities we get `ds = 25 kg * (1.12 * 10^4 m/s)^2 / (2 * 2000 amps * .50 Tesla * .5 meters) = 3.2 * 10^6 meters, or about 3200 km. **
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RESPONSE --> I did this differently and got a very different answer.
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16:30:01 query 28.66 u quark + 2/3 e and d quark -1/3 e counterclockwise, clockwise in neutron (r = 1.20 * 10^-15 m) What are the current and the magnetic moment produced by the u quark?
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RESPONSE --> My problem 66 does not seem to match this. I have a wire of length L and need to find the B field.
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16:30:05 ** If r is the radius of the orbit and v the velocity then the frequency of an orbit is f = v / (2 pi r). The frequency tells you how many times the charge passes a given point per unit of time. If the charge is q then the current must therefore be }I = q f = q v / (2 pi r). Half the magnetic moment is due to the u quark, which carries charge equal and opposite to the combined charge of both d quarks, the other half to the d quarks (which circulate, according to this model, in the opposite direction with the same radius so that the two d quarks contribute current equal to, and of the same sign, as the u quark). The area enclosed by the path is pi r^2, so that the magnetic moment of a quark is I A = q v / (2 pi r) * pi r^2 = q v r / 2. The total magnetic moment is therefore 2/3 e * v r / 2 + 2 ( 1/3 e * v r / 2) = 4/3 e v r / 2 = 2/3 e v r.. Setting this equal to the observed magnetic moment mu we have 2/3 e v r = mu so that v = 3/2 mu / (e r) = 3/2 * 9.66 * 10^-27 A m^2 / (1.6 * 10^-19 C * 1.20 * 10^-15 m) = 7.5 * 10^7 m/s, approx.. Note that units are A m^2 / (C m) = C / s * m^2 / (C m) = m / s. **
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16:31:35 query univ 28.68 (29.56 10th edition) infinite L-shaped conductor toward left and downward. Point a units to right of L along line of current from left. Current I. What is the magnetic field at the specified point?
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RESPONSE --> The horizontal wire has a magnetic field of 0. The vertical current provides the magnetic field of HALF of an infinite wire. B = 1/2 (micro I / 2*pi*r) short This is out of the page,.
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16:32:24 STUDENT RESPONSE FOLLOWED BY SOLUTION: I could not figure out the magnetic field affecting point P. the current is cursing ** I assume you mean 'coursing', though the slip is understandable ** toward P then suddnely turns down at a right angle. If I assume that the magnetic field of a thin wire is radial in all directions perpendicular to the wire, then it is possible that at least one field line would be a straight line from the wire to point P. It seems to me that from that field line,down the to the lower length of the wire, would affect at P. SOLUTION: The r vector from any segment along the horizontal section of the wire would be parallel to the current segment, so sin(theta) would be 0 and the contribution `dB = k ' I `dL / r^2 sin(theta) would be zero. So the horizontal section contributes no current at the point. Let the y axis be directed upward with its origin at the 'bend'. Then a segment of length `dy at position y will lie at distance r = sqrt(y^2 + a^2) from the point and the sine of the angle from the r vector to the point is a / sqrt(y^2 + a^2). The field resulting from this segment is therefore `dB = k ' I `dy / (a^2 + y^2) * a / sqrt(a^2 + y^2). Crossing the I `dy vector with the r vector tells us the `dB is coming at us out of the paper (fingers extended along neg y axis, ready to 'turn' toward r results in thumb pointing up toward us away from the paper). This is the direction for all `dB contributions so B will have the same direction. Summing all contributions we have sum(k ' I `dy / (a^2 + y^2) * a / sqrt(a^2 + y^2), y from 0 to -infinity). Taking the limit as `dy -> 0 we get the integral of k ' I a / (a^2 + y^2)^(3/2) with respect to y, with y from 0 to -infinity. This integral is -k ' I / a. So the field is B = - k ' I / a, directed upward out of the page. **
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RESPONSE --> I really was not sure how to calculate it further. I'm still confused after reading this.
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Ǟ]] assignment #031 031. `Query 31 Physics II 10-26-2008
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16:33:14 Query Principles and General Physics 21.04. A circular loop of diameter 9.6 cm in a 1.10 T field perpendicular to the plane of the loop; loop is removed in .15 s. What is the induced EMF?
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16:33:15 The average induced emf is the average rate of change of the magnetic flux with respect to clock time. The initial magnetic flux through this loop is flux = magnetic field * area = 1.10 T * (pi * .048 m)^2 = .00796 T m^2. The flux is reduced to 0 when the loop is removed from the field, so the change in flux has magnitude .0080 T m^2. The rate of change of
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16:33:17 flux is therefore .0080 T m^2 / (.15 sec) = .053 T m^2 / sec = .053 volts.
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16:33:20 query gen problem 21.23 320-loop square coil 21 cm on a side, .65 T mag field. How fast to produce peak 120-v output? How many cycles per second are required to produce a 120-volt output, and how did you get your result?
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16:33:23 The average magnitude of the output is peak output/sqrt(2) . We find the average output as ave rate of flux change. The area of a single coil is (21 cm)^2 = (.21 m)^2 and the magnetic field is .65 Tesla; there are 320 coils. When the plane of the coil is perpendicular to the field we get the maximum flux of fluxMax = .65 T * (.21 m)^2 * 320 = 19.2 T m^2. The flux will decrease to zero in 1/4 cycle. Letting t_cycle stand for the time of a complete cycle we have ave magnitude of field = magnitude of change in flux / change in t = 9.17T m^2 / (1/4 t_cycle) = 36.7 T m^2 / t_cycle. If peak output is 120 volts the ave voltage is 120 V / sqrt(2) so we have 36.7 T m^2 / t_cycle = 120 V / sqrt(2). We easily solve for t_cycle to obtain t_cycle = 36.7 T m^2 / (120 V / sqrt(2) ) = .432 second.+ A purely symbolic solution uses maximum flux = n * B * A average voltage = V_peak / sqrt(2), where V_peak is the peak voltage giving us ave rate of change of flux = average voltage so that n B * A / (1/4 t_cycle) = V_peak / sqrt(2), which we solve for t_cycle to get t_cycle = 4 n B A * sqrt(2) / V_peak = 4 * 320 * .65 T * (.21 m)^2 * sqrt(2) / (120 V) = .432 second.
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16:37:07 univ query 29.54 (30.36 10th edition) univ upward current I in wire, increasing at rate di/dt. Loop of height L, vert sides at dist a and b from wire. When the current is I what is the magnitude of B at distance r from the wire and what is the magnetic flux through a strip at this position having width `dr?
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RESPONSE --> Part A) Where I = i B = micro^2 / 2pi * r and into the page Part B) We need the differential in flux. BdA = micro^2 / 2pi*r Ldr Part C) Flux is the integral of d(flux) from a to b So we can pull ouyt micro IL / 2pi and inside the integral of dr / r from a to be. The integral of that would have the logarithmic function. So we would have micro I L / 2 pi * ln(b/a) Part D) E = dflux / dt = micro L / 2pi * ln(b/a) di/dt Part E) E = micro (0.240m / 2pi) * ln(0.360 /0.120) ( 9.60 A/s) = 5.06 x 10^-7 V
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16:37:53 ** The magnetic field due to the wire at distance r is 2 k ' I / r. The field is radial around the wire and so by the right-hand rule (thumb in direction of current, fingers point in direction of field) is downward into the page. The area of the strip is L * `dr. The magnetic flux thru the strip is therefore 2 k ' I / r * (L `dr). The total magnetic field over a series of such strips partitioning the area is thus sum(2 k ' I / r * L `dr, r from a to b). Taking the limit as `dr -> 0 we get } integral (2 k ' I / r * L with respect to r, r from a to b). Our antiderivative is 2 k ' I ln | r | * L; the definite integral therefore comes out to flux = 2 k ' L ln | b / a | * I. If I is changing then we have rate of change of flux = 2 k ' L ln | b / a | * dI/dt. This is the induced emf through a single turn. You can easily substitute a = 12.0 cm = .12 m, b = 36.0 cm = .36 m, L = 24.0 cm = .24 m and di/dt = 9.60 A / s, and multiply by the number of turns. **
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RESPONSE --> I know what I did was on the right track but whether it was all correct or not, I am not sure,
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