#$&* course Phy 242 7/21 1 030. `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: Force on a current = I * L * B sin(theta) = 150 amps * 160 m * 5.5 e^-5 T sin(65) = 1.2 N 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: F = q * v * B F = 1.6 e^-19 C * 8.75 e^5 m/s * .75 T = 1.05 e^-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):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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: N/A 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): N/A ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery univ 27.60 / 27.56 11th edition 27.60 (28.46 10th edition). cyclotron 3.5 T field. What is the radius of orbit for a proton with kinetic energy 2.7 MeV? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: From physics I, centripetal force = F = mv^2/r Magnetic field perpendicular to v = F = q v B We can set these equal: m v^2 / r = q v B Solving for r: r = mv / (qB) The kinetic energy of the proton = 2.7 e^6 * (1.6 e^-19 ) = 4.32 e^-13 J v = sqrt(2 KE /m) v = sqrt( 2 * 4.32 e^-13 J / 1.67 e^-27 kg) v = 2.3 e^7 m/s r = m v / (qB) = 1.67 e^-27 kg * 2.3 e^7 m/s / (1.6 e^-19 C * 3.5 T) r = .067 m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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) = 4.32 * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Good, however the solution has a wrong number where v = sqrt(2 KE / m) = sqrt(2 * 3.2 * 10^-13 J / (1.67 * 10^-27 kg) ) = 2.3 * 10^7 m/s 3.2 * 10^-13 J should be 4.32 * 10^-13 J everything else is correct though, it must have been entered wrong ------------------------------------------------ Self-critique Rating:ok
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Given Solution: ** 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. ** STUDENT QUESTION what numbers were used to find this? INSTRUCTOR RESPONSE In Problem 27.60, above, we found the radius of orbit for a proton with kinetic energy 2.7 MeV. Here we are finding the radius for a proton with twice the KE. We could do this in the same manner as before, and we would get the same result. However thinking in terms of the proportionality, as is done here, is both more efficient and more instructive. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok I used exact value for sqrt(2) so I got .095 instead of .096 ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qquery univ 27.74 / 27.72 11th edition 28.66 (was 28.52) rail gun bar mass m with current I across rails, magnetic field B perpendicular to loop formed by bars and rails What is the expression for the magnitude of the force on the bar, and what is the direction of the force? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: F = I L B F = ma a = I L B / m v_ave = (v0 + vf) / 2 = `ds / `dt a_ave = `dv / `dt = (vf - v0) / `dt vf ^2 = v0^2 + 2 a 'ds ds = (vf^2 - v0^2) / (2a) ds = vf^2 / 2a 'ds = v^2 / ((2 I L B )/ m) 'ds = m v^2 / (2 I L B) 'ds = 25 kg * (1.12 e^4 m/s)^2 / (2 * 2000 A * .5 T * .5 m) = 3.2 e^6 m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** STUDENT QUESTION I could find the magnitude of the force, but I don’t really understand the velocity equation laid out above. INSTRUCTOR RESPONSE vf^2 = v0^2 + 2 a `ds is one of the basic equations of uniformly accelerated motion, easily derived from the definitions of velocity and acceleration: vAve = (v0 + vf) / 2 = `ds / `dt aAve = `dv / `dt = (vf - v0) / `dt. We get the equations `ds / `dt = (v0 + vf) / 2 a = (vf - v0) / `dt from which we can eliminate `dt, obtaining vf^2 = v0^2 + 2 a `ds. Here, `ds stands for the displacement, v0 and vf for initial and final velocities, and a for acceleration on the interval in question. Armed with this equation, then, and having found the expression for a, knowing that the initial velocity is zero we easily solve for the `ds necessary to achive the desired vf. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qquery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: f = v / 2 pi r I = qf I = qv/(2 pi r) magnetic moment, mu = IA IA = q v ( 2 pi r) * pi r^2 IA = q v r / 2 Therefore the total moment = 2/3 e * v r/2 + 2(1/3 e * v r/2) = 2/3 e v r v = 3/2 mu /(e r) v = 3/2 * 9.66 e^-27 A m^2 / (1.6 e^-19 C * 1.2 e^-16 m) v = 7.5 e^7 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If r is the radius of the orbit and v the velocity then the frequency of an orbit is f = v / (2 pi r). The frequency tells you how many times the charge passes a given point per unit of time. If the charge is q then the current must therefore be I = q f = q v / (2 pi r). Half the magnetic moment is due to the u quark, which carries charge equal and opposite to the combined charge of both d quarks, the other half to the d quarks (which circulate, according to this model, in the opposite direction with the same radius so that the two d quarks contribute current equal to, and of the same sign, as the u quark). The area enclosed by the path is pi r^2, so that the magnetic moment of a quark is I A = q v / (2 pi r) * pi r^2 = q v r / 2. The total magnetic moment is therefore 2/3 e * v r / 2 + 2 ( 1/3 e * v r / 2) = 4/3 e v r / 2 = 2/3 e v r.. Setting this equal to the observed magnetic moment mu we have 2/3 e v r = mu so that v = 3/2 mu / (e r) = 3/2 * 9.66 * 10^-27 A m^2 / (1.6 * 10^-19 C * 1.20 * 10^-15 m) = 7.5 * 10^7 m/s, approx.. Note that units are A m^2 / (C m) = C / s * m^2 / (C m) = m / s. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qquery univ 28.78 / 28.70 11th edition 28.68 (29.56 10th edition) infinite L-shaped conductor toward left and downward. Point a units to right of L along line of current from left. Current I. What is the magnetic field at the specified point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: One side of the L shaped conductor is parallel to the Point. So the magnetic field due to that side would be 0. I am not really sure how to finish this problem... confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 'dB = k ' I ' dL / r^2 sin(theta) = 0 when parallel to current segment a segment of length 'dy at position y would be at distance r = sqrt(y^2 + a^2) using pythagoreans theorem from P the sin of the angle from the r vector to P is a/sqrt(qrty^2 + a^2) Therefore the field from this segment = 'dB = k ' I ' dy / (a^2 + y^2) * a / sqrt(a^2 + y^2) I 'dy X r comes out of the paper using the right hand rule this is the direction for all 'dB contributions so B will also have this direction sum(k ' I `dy / (a^2 + y^2) * a / sqrt(a^2 + y^2), y from 0 to -infinity). Take the limit as 'dy goes to 0 = integral k ' I a / (a^2 + y^2)^(3/2) with respect to y, with y from 0 to -infinity This integral = -k ' I / a Therefore the field = B = -k ' I / a ; directed upward out of the page ------------------------------------------------ Self-critique Rating:ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!