Assign_30_31

course Phy 202

Good work on questions and/or self-critiques.

Let me know if there's anything you need me to explain further.

]ĕ圯N Physics II 11-15-2005

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18:41:09 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 --> A charged particle passing through a combined electric and magnetic field, with particle velocity, electric field and magnetic field mutually perpendicular, will experience equal and opposite forces from the two fields and will therefore pass through the fields undeflected. I think this is what you are looking for.

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18:42:22 ** 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, i was a little off, i was looking more on the direction rather the force, but i see that the effect of an electric field does not depend on velocity

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18:43:32 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 --> If the particle goes in a perpendicular direction, then we could say that a force is exerted by that field to the particle

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18:43:43 ** 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 --> Exactly

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18:44:19 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 --> I am a little unsure about this, i am going to see what the solution provides

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18:48:34 ** 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, so at low enough velocities the magnetic force F = q v B is smaller in magnitude than the electrostatic force F = q E and at the high enough velocities the magnetic force is greater in magnitude than the electrostatic force. Also 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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18:48:59 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 --> What therefore is the current produced by the electron? v = sqrt(k*q1*q2 / (m r)) = 2.2 * 10^6

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18:50:27 **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 --> Ok so i got the velocity and you would use I = `dq / `dt = q / `dt, where q is the charge of the electron for finding the current and use B = 2 pi k ' I / r

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18:50:30 query univ 27.60 (28.46 10th edition). cyclotron 3.5 T field.

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18:50:32

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18:50:34 What is the radius of orbit for a proton with kinetic energy 2.7 MeV?

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18:50:36 ** 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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18:50:39 What is the radius of orbit for a proton with kinetic energy 5.4 MeV?

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18:50:41 ** 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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18:50:56 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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18:51:05 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 --> OK

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18:51:07 ** 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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18:51:10

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18:51:12 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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18:51:14 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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~e~ǂI} assignment #031 ]ĕ圯N Physics II 11-15-2005

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18:53:24 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 --> How fast to produce peak 120-v output? First you would find the area: (21 cm)^2 = (.21 m)^2 fluxmax would be .65 T * (.21 m)^2 * 720 How many cycles per second are required to produce a 120-volt output, and how did you get your result? THis i am not too sure on

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18:53:33 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 --> This is what i did for the beginning but i see what i did wrong in the end, i didn't know how the flux would be affected when it came to the cycle

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18:53:45 *&*& verify numbers using DERIVE and maybe change to symbolic approach *&*&

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18:53:49 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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18:53:51 ** 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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