#$&* course PHY 202 030. `Query 30
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Given Solution: ** 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. ** 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. ** 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. ** 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. ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qQuery 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The force on a current is found by using the equation: F= I * L * B sin(theta) = 150 amps * 160 meters * 5.5 * 10^-5 T * sin(65 deg) = 1.20 Newtons. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qQuery 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: We begin this problem by finding the force on an electron by using the equation: F= q v B With B being the magnetic field in teslas. = 1.6 * 10^-19 C * 8.75 * 10^5 m/s * .75 T = 1.05 * 10^-13 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Forgot about the vector direction. Only found the magnitude. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: If we know the orbital velocity of the electron, we can find the time it takes to go a certain distance. We know that the charge is 1 electron, so we can then find the charge/ time which is equal to the current. Given the current, we can then use our equation: B = k * 2 pi r I / r^2 In order to find the magnetic field B produced. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. ** STUDENT COMMENT I follow the part where you find the time required to travel around the circle and how to find the current. The velocity part is still a little hard for me to get. I just don't think I couldv'e came up with that formula on my own. Now that I see it it makes more sense though. Then the magnetic field is B = k*2pi*r*I/r^2 INSTRUCTOR RESPONSE Just as for orbiting satellites, the centripetal force is m v^2 / r. This time the force comes from Coulomb's Law, not Newton's Law of Universal Gravitation. However the process is otherwise the same: set centripetal acceleration equal to Coulomb force and solve for v. The fundamental relationship that governs magnetic fields created by currents is the following, which you should know: A current element I * `dL produces a magnetic field k ' I * `dL / r^2, at any point such that a line from that point to the current element is perpendicular to the current element. This condition applies to the magnetic field produced at the center by any short segment of a circular current. All the current segments around the circle have a total length of 2 pi r, so when we add up all the k ' * I * `dL / r^2 contributions of all the segments we get k ' * I * (2 pi r) / r^2. (This expression can of course be simplified, but leaving it in its 'raw' form for now emphasizes the use of the circumference 2 pi r.) So basically, because of the symmetry of the circle about its center, we can replace `dL by the total circumference 2 pi r. The result is the magnetic field at the center of the loop. The formula, in case you prefer to memorize it, is B = 2 pi k ' I / r. It's fine to memorize this formula, but an alternative to cluttering up your memory is to understand how it is obtained by replacing `dL in the basic law with the circumference 2 pi r. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!