qj assignment #024 rYɐZ Physics II 05-06-2006
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13:04:14 Query problem set 3 #'2 1-6. How do we determine the current in the circuit and the voltage across each resistor when we know the voltage across a series combination of two known resistances?
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RESPONSE --> To find current we use the formula I= V/R and we add resistances to get R.
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13:04:59 ** To get the current calculate I = V / R, where R is the sum of the two resistances. To get the voltage across each resistor calculate V = I * R for each resistor. **
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RESPONSE --> ok, to find voltage just rearrange the formula v=I*R
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13:06:57 How do we determine the current and voltage across each resistor when we know the voltage across a parallel combination of two known resistances?
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RESPONSE --> We determine the current by dividing the voltage by the resistance for each resistor I=V/R We determine the voltage by mulitplyig the current by the resistances for each resistor. V=I*R WE add the two currents to get the total current..
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13:08:13 ** The voltage across both resistors is the same and is equal to the voltage across the combination. The current in each resistor is calculated by I = V / R. The total current is the sum of the two currents. **
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RESPONSE --> ok, the total current is the sum of the two currents
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13:12:01 A series circuit contains a capacitor of known capacitance and a resistor of known resistance. The capacitor was originally uncharged before the source voltage was applied, and is in the process of being charged by the source. If we know the charge on the capacitor, how do we find the current through circuit?
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RESPONSE --> If charge (Q) is knows on the capacitor we can find the voltage throught the capacior witht the formula V=Q/C If it discharges through a resistance R, then current will flow at the rate I = Vcap / R = Q / (R C).
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13:13:14 The voltage across the capacitor is equal to the charge divided by capacitance. The voltage across the capacitor opposes the voltage of the source. Since the voltage drop around the complete circuit must be zero, the voltage across the resistor is the difference between source voltage and the voltage across the capacitor. Dividing the voltage across the resistor by the resistance we obtain the current thru the circuit.
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RESPONSE --> ok
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13:17:09 If we know the capacitance and initial charge on a capacitor in series with a resistor of known resistance then how to we find the approximate time required for the capacitor to discharge 1% of its charge through the circuit?
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RESPONSE --> If this capacitor holds a charge Q, then we find voltage Vcap with the formula C = Q / Vcap Vcap = Q / C. When the charge is built by the current from the source, its voltage will resist that of the source, since the positive terminal of the source tends to move positive charges to the plate of the capacitor to which it is most directly connected and the negative terminal to the plate connected to it. So if the voltage of the source is Vs, the voltage across the resistor will be Vr = Vs - Vcap. It follows that the current through the resistor is I = (Vs - Vcap) / R. WE then take 1 percent of the initial charge (q*.1) and divide it by the current to get the time.
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13:17:42 ** From capacitance and initial charge we find the voltage. From the voltage and the resistance we find the current. We take 1% of the initial charge and divide it by the current to get the approximate time required to discharge 1% of the charge. } This result is a slight underestimate of the time required since as the capacitor discharges the current decreases. **
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RESPONSE --> ok, this is a underestimate because as the capacitor charge decreases so does the current.
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W߲}lVv assignment #025 rYɐZ Physics II 05-06-2006
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13:41:23 Query Gen Phy 16.15 charges 6 microC on diagonal corners, -6 microC on other diagonal corners of 1 m square; force on each. What is the magnitude and direction of the force on the positive charge at the lower left-hand corner?
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RESPONSE --> To determine this we have to use Coulombs law. F=kq1q2/r^2 F=9*10^9(6*10*-6)(6*10*-6)/(.1m)^2=3.24N For the diagnol we take the square root of .2=.44 We then plug into the above formula F=9*10^9(6*10*-6)(6*10*-6)/(.44m)=.736N Therefore to the right and left there is a force of 3.24N Diagnally there is a force of .736N
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13:44:09 ** The charges which lie 1 meter apart are unlike and therefore exert attractive forces; these forces are each .324 Newtons. This is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6 * 10^-6 C) / ( 1 m)^2 = 324 * 10^-3 N = .324 N. Charges across a diagonal are like and separated by `sqrt(2) meters = 1.414 meters, approx, and exert repulsive forces of .162 Newtons. This repulsive force is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6 * 10^-6 C) / ( 1.414 m)^2 = 162 * 10^-3 N = .162 N. The charge at the lower left-hand corner therefore experiences a force of .324 Newtons to the right, a force of .324 Newtons straight upward and a force of .162 Newtons at 45 deg down and to the left (at angle 225 deg with respect to the standard positive x axis, which we take as directed toward the right). This latter force has components Fy = .162 N sin(225 deg) = -.115 N, approx, and Fx = .162 N cos(225 deg) = -.115 N. The total force in the x direction is therefore -.115 N + .324 N = .21 N, approx; the total force in the y direction is -.115 N + .324 N = .21 N, approx. Thus the net force has magnitude `sqrt( (.21 N)^2 + (.21 N)^2) = .29 N at an angle of tan^-1( .21 N / .21 N) = tan^-1(1) = 45 deg. The magnitude and direction of the force on the negative charge at the lower right-hand corner is obtained by a similar analysis, which would show that this charge experiences forces of .324 N to the left, .324 N straight up, and .162 N down and to the right. The net force is found by standard vector methods to be about .29 N up and to the left. **
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RESPONSE --> I used .1m instead of 1m. However, it looks like I got the steps right up until the force magnitude was calculated. This latter force has components Fy = .162N sin(225 deg) =-.115 N and Fx =.162 Ncos(225 deg) = -.115 N. WE then sum the two to get the total force. The total force in the x direction is -.115 N + .324 N = .21 N. The total force in the y direction is -.115 N + .324 N = .21 N Thus, the net force has magnitude `sqrt( (.21 N)^2 + (.21 N)^2) = .29 N at an angle of tan^-1( .21 N / .21 N) = tan^-1(1) = 45 deg. The magnitude and direction of the force on the negative charge at the lower right-hand corner is obtained by a similar analysis, which would show that this charge experiences forces of .324 N to the left, .324 N straight up, and .162 N down and to the right. The net force is found by standard vector methods to be about .29 N up and to the left. **
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13:44:15 query univ 21.68 (22.52 10th edition) 5 nC at the origin, -2 nC at (4 cm, 0). If 6 nC are placed at (4cm, 3cm), what are the components of the resulting force?
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RESPONSE --> ok
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13:44:19 ** The 5 nC charge lies at distance sqrt( 4^2 + 3^2) cm from the 6 nC charge, which is 3 cm from the charge at (4 cm ,0). The force exerted on the 6 nC charge by the two respective forces have magnitudes .00011 N and .00012 N respectively. The x and y components of the force exerted by the charge at the origin are in the proportions 4/5 and 3/5 to the total charge, giving respective components .000086 N and .000065 N. The force exerted by the charge at (4 cm, 0) is in the negative y direction. So the y component of the resultant are .000065 N - .00012 N = -.000055 N and its x component is .000086 N. **
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RESPONSE --> ok
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13:44:23 Query univ 21.76 (10th edition 22.60) quadrupole (q at (0,a), (0, -a), -2q at origin). For y > a what is the magnitude and direction of the electric field at (0, y)?
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RESPONSE --> ok
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13:44:26 ** The magnitude of the field due to the charge at a point is k q / r^2. For a point at coordinate y on the y axis, for y > a, we have distances r = y-a, y+a and y. The charges at these distances are respectively q, q and -2q. So the field is k*q/(y - a)^2 + k*q/(y + a)^2 - 2k*q/y^2 = 2*k*q*(y^2 + a^2)/((y + a)^2*(y - a)^2) - 2*k*q/y^2 = 2*k*q* [(y^2 + a^2)* y^2 - (y+a)^2 ( y-a)^2) ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - (y^2 - a^2)^2 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - y^4 + 2 y^2 a^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [ 3 a^2 y^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) . For large y the denominator is close to y^6 and the a^4 in the numerator is insignifant compared to a^2 y^2 so the expression becomes 6 k q a^2 / y^4, which is inversely proportional to y^4. **
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RESPONSE --> ok
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13:44:30 query univ 22.102 annulus in yz plane inner radius R1 outer R2, charge density `sigma.What is a total electric charge on the annulus?
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RESPONSE --> ok
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13:44:34 ** The total charge on the annulus is the product Q = sigma * A = sigma * (pi R2^2 - pi R1^2). To find the field at distance x along the x axis, due to the charge in the annulus, we first find the field due to a thin ring of charge: The charge in a thin ring of radius r and ring thickness `dr is the product `dQ = 2 pi r `dr * sigma of ring area and area density. From any small segment of this ring the electric field at a point of the x axis would be directed at angle arctan( r / x ) with respect to the x axis. By either formal trigonometry or similar triangles we see that the component parallel to the x axis will be in the proportion x / sqrt(x^2 + r^2) to the magnitude of the field from this small segment. By symmetry only the x component of the field will remain when we sum over the entire ring. So the field due to the ring will be in the same proportion to the expression k `dQ / (x^2 + r^2). Thus the field due to this thin ring will be magnitude of field due to thin ring: k `dQ / (x^2 + r^2) * x / sqrt (x^2 + r^2) = 2 pi k r `dr * x / (x^2 + r^2)^(3/2). Summing over all such thin rings, which run from r = R1 to r = R2, we obtain the integral magnitude of field = integral ( 2 pi k r x / (x^2 + r^2)^(3/2) with respect to r, from R1 to R2). Evaluating the integral we find that magnitude of field = 2* pi k *x* | 1 /sqrt(x^2 + r1^2) - 1 / sqrt(x^2 + r2^2) | The direction of the field is along the x axis. If the point is close to the origin then x is close to 0 and x / sqrt(x^2 + r^2) is approximately equal to x / r, for any r much larger than x. This is because the derivative of x / sqrt(x^2 + r^2) with respect to x is r^2 / (x^2+r^2)^(3/2), which for x = 0 is just 1/r, having no x dependence. So at small displacement `dx from the origin the field strength will just be some constant multiple of `dx. **
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RESPONSE --> ok
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SŎ[o}ꆉ assignment #026 rYɐZ Physics II 05-06-2006
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14:08:40 Query 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 --> WE have to use the formula E=kq/r^2 Since we are in the middle and there is a total distance of .16 meters we use a r=.08m Also since we are in the middle the field of 745N/C=745/2=372.5n/c We then use the above formula 372.5=9*10^9*Q/.08^2 2.384=9*10^9q 2.65*10^-10=Q
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14:10:51 ** The charges are each 8 cm from the field point. 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) = 372.5 N/C. Thus E = 372.5 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 = 372.5 N/C * (.08 m)^2 / (9 * 10^9 N m^2 / C^2) = 372.5 N/C * .0064 m^2 / (9 * 10^9 N m^2 / C^2) = 2.6 * 10^-10 C, approx. **
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RESPONSE --> ok
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14:12:19 If the charges are represented by Q and -Q, what is the electric field at the midpoint?
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RESPONSE --> at the midpoint (.08m) the charges are Q and -Q the elecric field would be E=2KQ/R^2. This is because we have two Q's
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14:12:52 ** 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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RESPONSE --> ok
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14:15:53 Query electrostatics. Explain why a charge enclosed in an aluminum cylinder is not affected by a charge outside the aluminum cylinder, when the charges would clearly affect one another in the absence of the cylinder.
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RESPONSE --> The aluminum can is full of electrons moving around freely. As the outside electrons attract those in the cylinder they are able to move inside the cylinder so the field cannot get inside the cylinder.
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14:17:23 STUDENT RESPONSE AND INSTRUCTOR COMMENT: The electrons in a conductor will freely move and redidtribute them selves so that the field can not penetrate the cylinder. INSTRUCTOR COMMENT: Good explanation. See the next paragraph for a more detailed explanation: the aluminum cylinder is full of free charges that can migrate freely anywhere on the cylinder. The outside charge attracts opposite charges, which build to greatest density on regions of the aluminum cylinder nearest it. This process must continue until there is no electric field in the aluminum, because if there is an electric field in the aluminum then charges will move in response to it. Movement of charges must continue until the field is eliminated. This effectively 'shields' all point inside the cylinder from the effect of external charges.
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RESPONSE --> ok,the aluminum cylinder is full of free charges that can flow freely anywhere on the cylinder. The outside charge attracts opposite charges, which build to greatest density on regions of the aluminum cylinder nearest it. This process must continue until there is no electric field in the aluminum, because if there is an electric field in the aluminum then charges will move in response to it. Movement of charges must continue until the field is eliminated. This effectively 'shields' all point inside the cylinder from the effect of external charges.
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14:19:57 Explain why bringing a charged plastic rod near the far end of an aluminum rod will have a direct affect on a charge close to the near end of the rod, whereas the same charge and plastic rod will exhibit no measurable interaction in the absence of the aluminum rod.
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RESPONSE --> Because opposites charges will attract between the two rods. When they are brought togeteher will change the charges so they are opposite.
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14:20:41 ** This question concerns the situation where a charged object is located near one end of the metal rod and the charged plastic rod is brought near the other end of the metal rod. Nothing ever touches anything else so there is no migration of charge from one object to the other. When the charged plastic rod is brought near there is however a redistribution of charge on the conducting rod and the force on the charged object will change. The redistribution of charges causes the end of the pipe near the rod to take on a charge opposite to that of the rod; this occurs by displacement of charge from the other end of the rod, which therefore ends up with the same type of charge as the plastic rod. Thus the charge near that end will experience a force in the same direction as if the rod was near. If the metal rod wasn't there, the distance of the plastic rod would cause the effect on that charge to be minimal. **
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RESPONSE --> ok, there will be a redistribution of charges at the end of the pipe near the rod to take on a charge opposite to that of the rod; this occurs by displacement of charge from the other end of the rod, which therefore ends up with the same type of charge as the plastic rod. Thus the charge near that end will experience a force in the same direction as if the rod was near. If the metal rod wasn't there, the distance of the plastic rod would cause the effect on that charge to be minimal. **
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14:20:47 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 --> ok
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14:20:50 ** Flux is not a vector quantity so your flux values will not be multiples of the i, j and k vectors. The normal vectors 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 --> ok
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14:20:58 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 --> ok
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14:21:02 ** 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 --> ok
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14:21:06 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 --> ok
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14:21:09 ** Gaussian surfaces for this configuration are cylinders of radius r and 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. For r < a the charge enclosed by the Gaussian surface is L * alpha so that the flux is 4 pi k L * alpha and the electric field is 4 pi k L * alpha / (2 pi r L ) = 2 k alpha / r. For a < r < b the Gaussian surface lies within the conductor so the field is zero. This implies that 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 charge -L * alpha, so that the inner surface is characterized by 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 density includes that of the line charge as well as the inner surface of the shell the outer surface of the shell has charge density 2 alpha. **
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RESPONSE --> ok
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ȌͰoƼہ assignment #027 rYɐZ Physics II 05-06-2006
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14:39:41 Note that my solutions 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 adapt the two notations, since k = 1 / (4 pi epsilon0), or alternatively epsilon0 = 1 / (4 pi k).
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RESPONSE --> ok
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14:41:44 Query RC circuits. Explain why the current in a discharging RC circuit, consisting of an initially charged capacitor discharging through a resistor, decreases exponentially with time.
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RESPONSE --> The current in a discharging RC circuit, consisting of an initially charged capacitor discharging through a resistor, decreases exponentially with time because the voltage decreased and voltage is related to current. As voltage decreases so does the current.
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14:42:51 ** The exponential decrease results because the current is proportional to the voltage, which is proportional to the charge. The current is the rate at which charge changes. Thus the rate at which the charge changes is proportional to the charge. Whenever the rate at which something changes is proportional to that thing (e.g., population of a species in a noncompeting environment, money in an interest-bearing account) the change will be exponential. **
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RESPONSE --> ok, Whenever the rate at which something changes is proportional to something the change will be exponential.
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14:45:01 Explain why the current in a charging RC circuit, consisting of an initially uncharged capacitor in parallel with a resistor, decreases exponentially with time.
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RESPONSE --> As the capacitor is charged less voltage passes through the capacitor. As the charge increases on the capacitor it has voltage that acts against the generator. So as charge increases on the capacitor the current decreases.
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14:45:27 STUDENT RESPONSE WITH INSTRUCTOR COMMENT: This may not be right. Electrons flow out form the negative terminal of the battery in the charging RC circuit, through the resistor and accumulate on the upper plate of the capacitor. The electrons will flow into the positive terminal which leaves a positive charge on the other plate of the capacitor. This charge accumulates on the capacitor the potential difference across it increases and the current decreases until the capacitor equals the emf of the battery. INSTRUCTOR COMMENT: This is close to a complete answer. The current flow increases the charge on the capacitor, which results in a voltage that opposes the source; the more charge the more opposition and the less current flow. **
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RESPONSE --> ok, as charge on the capacitor increases, the voltage on the capacitor opposes the source. This results in more charge, more opposition and less current flow.
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14:54:11 Query gen phy 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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RESPONSE --> A. We use the formula V=Q/R V=9.0*10^9(1.6*10^-19/2.5*10^-15 V=576,000V B. W=q(kQ/rb-KQ/ra) W=1.6*10^-19C*576000v=9.216*10^-14J
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14:54:41 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/C62(1.60*10^-19C)/2.5*10^-15m = 5.8*10^5V. Part B was the more difficult potrion of the problem. You have to consider a system that consists of two protons 2.5*10^-5m apart. The work done by the electric field is W = qV. Thus the potential energy will be equal to the work done by the system. PE=(1.60*10^-19C)(5.8*10^5V) =9.2*10^-14J.
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RESPONSE --> ok
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14:54:45 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 --> ok
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14:54:48 ** The voltage would be 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 --> ok
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14:54:50 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 --> ok
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14:54:53 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 --> ok
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yy͍Dܷ assignment #028 rYɐZ Physics II 05-06-2006
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21:26:48 Query magnetic fields produced by electric currents. What evidence do we have that electric currents produce magnetic fields?
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RESPONSE --> A radio or a generator. The magnetic force is what makes each work. We can also look at electrical appliances and see their EMF or electric magnetic fields. THis is evidence.
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21:27:37 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 --> Ok, a good example is 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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21:31:30 How does the direction of an electric current compare with the direction of the magnetic field that results?
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RESPONSE --> WE use the right hand rule. If the right hand is used as a model, and the thumb of the right hand points along the wire from positive towards the negative side (""conventional current"", the reverse of the direction of actual movement of electrons), then the magnetic field will wrap around the wire in the direction indicated by the fingers of the right hand
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21:32:25 ** 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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RESPONSE --> ok
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21:36:22 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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RESPONSE --> WE use the formula C=Eo A/D. The constant Eo is the permittivity of free space (8.85*10^-12 C^2N*m^2) WE then substitute into this formula .20F=8.85*10^-12 * A/2.2*10^-3 4.4*10^-4=8.85*10^-12 * A A=4.97*10^7m^2
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21:38:51 ** 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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RESPONSE --> I see the way you did it but the text book says to do it the way I did. Either way will work..correct?
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21:42:28 Query Gen Phy Problem Charge Q is placed on a parallel-plate capacitor with air between the plates; the separation of the plates is halved as a dielectric with const K inserted. By what factor does the energy stored on the capacitor change? Compare the new electric field with the old.
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RESPONSE --> If we decrease the distance of the plates by half we would cut the capacity in half C=E0 (.a/.5d) the capacitance doubles.
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21:43:26 ** 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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RESPONSE --> ok, I see. C will increase by factor k, and will also increase by factor 2 due to halving of the distance. 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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21:43:29 ......!!!!!!!!...................................
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RESPONSE --> ok
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21:52:23 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 --> WEll we use the formula B = k ' I L / r^2 * sin(theta) The magnitude of a magnetic field due to source I*L , at distance r from the source so the angle between I and a line from the source to the point is `theta.
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21:52:43 ** 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 --> ok
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21:53:44 Explain how to obtain the magnetic field due to a circular loop at the center of the loop.
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RESPONSE --> The loop of radius r can be thought of as a series of segments. The total length is 2 `pi r equal to the circumference of the circle. Each segment `dL is a source I `dL and is perpendicular to a line from the center of the segment to the center of the circle. Each segment thus contributes `dB = k ' I `dL / r^2 to the field at the center. All contributions are in the same direction. So we get the formula B = k ' I / r^2 * 2 `pi r = 2 `pi k ' I / r.
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21:54:48 ** 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 --> ok
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21:54:52 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 --> ok
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21:54:55 ** 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 store 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) The integration process becomes necessary for cylindrical and spherical capacitors and other capacitors which are not in a parallel-plate configuration. **
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RESPONSE --> ok
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21:54:58 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 thecapacitance, charge of each plate, electric field, energy stored?
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RESPONSE --> ok
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21:55:00 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 --> ok
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21:55:04 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 --> ok
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21:55:06 ** 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 --> ko
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ﮈyvxy assignment #029 rYɐZ Physics II 05-06-2006
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22:31:42 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 --> The flux of any vector field through an area is equal to the product of the field strength and the area, as long as the vector field is directed perpendicular to the area. flux = B * A If the field is directed at angle `theta from perpendicular, the flux will be reduced by factor cos(`theta). flux = B * A * cos(`theta).
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22:32:31 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 --> ok, strength of field in tesla.
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22:36:51 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 --> If it is oriented perpendicular to the magnetic flux will be: initial flux = B * A If the field is parallel to the plane, the flux will become 0 flux change = 0 - `phi = -`phi = - B * A. A flux change of -`phi in time interval `dt gives an average rate of flux change `d`phi / `dt = - (B * A) / `dt
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22:38:41 ** 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 --> ok
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22:43:54 Explain how alternating current is produced by rotating a coil of wire with respect to a uniform magnetic field.
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RESPONSE --> Simply moving the wire through the magnetic field causes electric current to flow in the wire. It will go perpendicular to parallel or highest energy to 0 as it spins.
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22:44:40 ** 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 --> ok, The changing magnetic flux produces voltage, which in turn produces current.
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22:51:38 Query problem of 18.39; compare power loss if 520 kW delivered at 50kV as opposed to 12 kV thru 3 ohm resistance
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RESPONSE --> Firs we have to find the Amps for each delivery 50kv and 12kv. P=IV 620000kw=I*12000 I=51.67A 620000=I*520000 I=12.4A Now that we know amps we have to factor in the resistance of 3 ohms V=IR V=12.4*3=37.2V V=51.67*3=155.01 With volts for each system we can find the power. P=IV P=12.4*37.2=461.28 P=155.01*51.67=8009.37 So we have a ratio of 461.28:8009.37
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22:54:57 ** 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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RESPONSE --> ok, we have different answers but I used 620000 wats instead of 520,000.
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22:55:01 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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RESPONSE --> ok
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22:55:03 ** 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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RESPONSE --> ok
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аۓcӯMT} assignment #030 rYɐZ Physics II 05-06-2006
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23:14:30 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 --> It is independent of velocity.
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23:14:40 ** 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 --> ok, There is a force due to the electric field between the plates. The plates of a capacitor do not create a magnetic field.
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23:15:17 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 --> Generally if v = E / B, the force exerted by the electric field on a charge q is q E and the force exerted by the magnetic field has magnitude q v B = E. If the fields are appropriately oriented the forces will be equal and opposite, and any charge q moving at this velocity will pass through the fields without a change in direction.
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23:16:20 ** 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 --> ok
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23:19:58 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 --> At high velocities magnetic is greater than electric. At relatively low the magnetic force is smaller than electric force.
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23:20:55 ** 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 --> ok, 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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23:30:04 Query Gen Phy Determine the magnetic field produced at the location of a proton by an electron in circular orbit at a distance of 5.29 * 10^-11 meters. Both electron and proton have charges of magnitude 1.6 * 10^-19 C, and electron mass is 9.11 * 10^-31 kg. What is the velocity of the electron (hint: the necessary centripetal force is provided by the Coulomb attraction)? What therefore is the current produced by the electron? What is the field produced by the electron at the nucleus? How did you calculate the magnetic field produced by this current?
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RESPONSE --> I'm not sure but I would think we would use the formula q v B / m = v^2 / r,
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23:31:47 **If you know the orbital velocity of the electron and orbital radius then you can find 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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RESPONSE --> I was way off. I was unsure of your hint. I see how you went about it. First find the velocity then the 'dt. With that we can find the current. And finally the magnetic field. It's like putting a puzzle together.
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23:31:51 query univ 27.60 (28.46 10th edition). cyclotron 3.5 T field.
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RESPONSE --> ok
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23:32:00 What is the radius of orbit for a proton with kinetic energy 2.7 MeV?
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RESPONSE --> ok
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23:32:04 ** 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 --> ok
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23:32:06 What is the radius of orbit for a proton with kinetic energy 5.4 MeV?
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RESPONSE --> ok
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23:32:08 ** 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 --> ok
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23:33:04 query 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 --> I wan't sure if we were supposed to do this for PHY 202. This problemw as in my book.
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23:33:13 ** 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 --> ok
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23:34:18 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 --> Again, I wasn't sure about this one.
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23:34:39 ** 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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RESPONSE --> ok
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23:34:48 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 --> ok
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23:34:51 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 --> ok
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ɟWLԐt݃Ѕ assignment #031 rYɐZ Physics II 05-07-2006
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00:13:17 query zhen Phy 21.23 720-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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RESPONSE --> We know the magnetic field is .65t the number of coils (720) and the heigh of the coil 21cm First we need ot find the area of the coil (.21m)^2=.0441 With this information we can fint the flux flux=.65*.0441* 720 =20.64 In a quarter of a cycle the flux will =0. S we can find the average with = 20.64/.25T=82.56T WE can find the average voltage with 120v/'sqrt 2=84.85 We can solve for t with 82.56T=84.85 t=1.0277
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00:17:01 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 720 coils. When the plane of the coil is perpendicular to the field we get the maximum flux of fluxMax = .65 T * (.21 m)^2 * 720 . The flux will decrease to zero in 1/4 cycle. Letting T stand for the time of a complete cycle we have ave magnitude of field = magnitude of change in flux / change in t = .65 T * .(21) m^2 * 720 / (1/4 T). If peak output is 120 volts the ave voltage is 120 V / sqrt(2) so we have .65 T * (.21 m)^2 * 720 / (1/4 T) = 120 V / sqrt(2). We easily solve for T to obtain T = .65 T * (.21 m)^2 * 720 * sqrt(2) / (120 V) =
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RESPONSE --> I was close. What happened to the 1/4T in your equation? would I only be solving for 1/4 of the time?
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00:17:11 *&*& verify numbers using DERIVE and maybe change to symbolic approach *&*&
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RESPONSE --> ok
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00:17:14 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 --> ok
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00:17:16 ** 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 --> ok
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