course PHY 202 assignment #009009. `Query 30
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20:25:27 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 --> The force due to the field between two capicitor plates is given by : F=q(E) where F is the force, q is the charge particle and E is the magnitude of the electric field. Thus the force in this situation will not be affected by its velocity.
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20:25:41 ** 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
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20:28:19 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 --> The force on a charged particle due to the magnetic field created by a wire is given by F=qvB, where q=charge, v=velocity, and B=strengh of the magnetic field. Thus the force will be affected by the velocity of the charged particle as it is multiplied by strength of the magnetic field and the charge to get the force.
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20:29:35 ** 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 --> Use F=qvb only if the particle moves in a direction perpendicular to the coil. If not, then must use multiply qvb by sin (theta) to get the force.
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20:37:46 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 --> The velocity in this situation is given by: E/B, where E is the strength of the electric field and B is the strength of the magnetic field. To get the net force, we must determine the force due to each of the fields. The force due to the electric field is given by F=qE, where q=charge and E=strength of the electric field. The force due to the magnetic field is given by F=qvB, were q=charge, v=velocity calculated above, and B=strength of the magnetic field. One will find that the magnitude of the force of the electrical field and the force due to the magnetic field will be equal. Since the force exerted by the magnetic field is perpendicular to the electric field and to the particle velocity,it must be in the direction of the electrical field or opposite. Thus, the two forces are equal in magnitude,,yet opposite in direction, so the particle will experience no net force.
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20:38:16 ** 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
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20:46:58 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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RESPONSE --> Given the formula: F=(I)(L)(B)sin 65. Where: I=Current, L=Length of wire, B=Strength of magnetic field Thus:F=(150)((160)(5.5E-5)(.906)=1.19N
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20:47:22 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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RESPONSE --> ok
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20:53:24 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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RESPONSE --> Charge of electron: 1.602E-19C Given the formula: F=absolute value of qvB, where q=charge, v=velocity, B=strength of magnetic field. So: F=(1.602E-19)(8.75E5)(.75)=1.05E-13. Using the right hand rule, I determine that the force is moving in the direction to the north.
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20:53:54 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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RESPONSE --> ok
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22:51:16 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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RESPONSE --> I don't think we were given enough info to solve this problem. I feel like I will need the mag of Force or strength of the magnetic force to solve it correctly. To find the velocity of the electron, I would use, the formula, r=mv/qB and solve for v. Where v is the velocity, r is the radius, q is the charge, and B is the strength of the magnetic field. Here, I think we are missing the strength of the magnetic field and cannot solve completely. But If I were given the strength of the magnetic field, I would solve for velocity. But now, I am not quite sure how I would get the current. If I had the strength of the magnetic field, I could use the formula B=4piE-7(I)/2pi(r) and solve for the current, but I am not quite sure where to go from here. It seems like I am missing something.
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23:08:33 **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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RESPONSE --> I was on the right track with using the Centripital force, I just used it incorrectly. It turns out that we do have enough info to solve the problem. I never would have thought to set the centripetal force =coulomb attraction for the orbital radius. With this equation, I only have one unknown, and that is v. K=9.0E9, q1=q2=1.602E-19, and r is given at .529E-10, and m=9.01E-31 kg. Current is actually charge/unit of time, amps or c/s. Then the rest, of the problem is pretty straightforward.
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J}cֳ^ assignment #009 009. `Query 30 Physics II 06-25-2007 ͉zT} assignment #009 009. `Query 30 Physics II 06-25-2007 J}cֳ^ assignment #009 009. `Query 30 Physics II 06-25-2007 ͉zT} assignment #009 009. `Query 30 Physics II 06-25-2007
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23:36:20 What is the radius of orbit for a proton with kinetic energy 2.7 MeV?
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RESPONSE --> I would use kinetic energy to solve for my v. Using the formula KE=1/2(m)(v). Where m is the mass of the proton and v is the unknown velocity. Thus, (v)=6.0E30. Then set centripetal force =coloumb attraction and solve for r. Thus, r=k(Q)/m(v^2) where, K is the prop. constant, q is the charge, m is the mass, and v is the velocity. Thus solving for r=4.41E81.
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23:41:50 ** 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 --> Solved this whole problem incorrectly! Primarily solved for v incorrectly and used a different formula to calculate the r. I am noticing now that the value of B was given. I think i read the problem too quickly and overlooked that. I'm sure I would have use r=mv/(qB) if I would I have noticed the value of B was given.
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瑚ٮl콍潈 assignment #009 009. `Query 30 Physics II 06-26-2007
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18:00:42 What is the radius of orbit for a proton with kinetic energy 5.4 MeV?
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RESPONSE --> First Ke is given as 5.4MeV= which equates to 5.4E6*1.6E-19=8.64E-13 Joules I can use the formula: r=mv^2/qB,where r is the radius, m is the mass of the proton, v is the velocity, q is the charge and B is the strength of the magentic field to solve v. v=sqrt of 2(KE)/m=sqrt(2*8.64E-13/1.67E-27)= 3.21E7 so...r=mv/qb=1.67*E-27*3.21E7/1.67E-19(3.5T)= .091 m
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18:01:48 ** 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 --> I could have easily arrived at the same answer by thouroughly understanding the relationship between the radius and the velocity, which are directly proportional to each other.
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~|JV^GȌP assignment #009 009. `Query 30 Physics II 06-26-2007"