course PHY 202 ??????k??????assignment #007
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21:44:15 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 --> The magnitude of the magnetic field due to a given current is given by: B=k'(IL)/r^2. If the position of the point with respect to the segment is at an angle then you would have B=k'(IL)/r^2(sin theta). The angle that a vector from the point with the respect to the segment makes is the angle that is multiplied by r^2.
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21:44:28 ** 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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21:47:09 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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RESPONSE --> CV=Q, where C is the Capitance, V is the Voltage and Q is the charge. 7.0E-6(12)=8.4E-5.
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21:49:29 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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RESPONSE --> My answer takes into account that microF is equivalent to 10E-6 F.
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????????????? assignment #007 007. `Query 28 Physics II 06-20-2007
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21:20:19 Query magnetic fields produced by electric currents. What evidence do we have that electric currents produce magnetic fields?
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RESPONSE --> The compass needle provides the evidentce that we need that electric currents produce magnetic fields. IF the needle is placed near a wire with an electric current, it will deflect. This deflection is caused by the force of the magnetic field.
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21:22:20 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 --> It's important to note that the compass will not only be deflected when placed near a wire with a current, but if it is originally oriented in the North South direction, the needle will be deflected towards the East-West direction. This is because the needle needs to align tangent to a circle around the wire.
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21:32:00 How is the direction of an electric current related to the direction of the magnetic field that results?
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RESPONSE --> Magnetic fields that are produced by an electric current are oriented in circles areound the wire at the center. However, the right hand rule must be used to determine the direction of the magnetic fields.
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21:32:34 ** 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, this answer went into detail about how to do the right hand rule.
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21:38:48 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 --> The area is given by the formula:CD/epsilon omicron. Where C=Capitance of plates D= Distance that plates are seperated. Epsilon Omicron=8.85E-12. .20F(.0022m)/8.85E-12= 4.9E7.
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21:39:10 ** 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 --> Units are m^2
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21:47:57 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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RESPONSE --> The energy storage will change by a factor of 2. Using the formula for energy storage which is PE=1/2(Q^2)/C IF the Capitance is halved, then the energy storage will increase by a factor of 2. I think the new electric field compared to the old one would increase as well.
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22:01:28 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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RESPONSE --> It's interesting to note that the electric field is independent of the distance between plates, so halving the distance will halve the distance between plates. However, according to the solution given, I think my answer was correct.
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???????Q???x~??assignment #007 007. `Query 28 Physics II 06-20-2007"