#$&* course phy 242 july 25 Question: `qQuery introductory problem set 54 #'s 14-18.
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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 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: Equation to be used: F= mv^2/r , F= q*B*v Now given the equations we eliminate to attain mv^2/r = q*B*v which then manipulate further to get r = mv/ (qB) Now we need to find the velocity for the equation so we use the equation v= sqrt(2*KE/m). v= sqrt(2*(4.33*10^-13J)/ (1.67*10^-27 kg))= 2.3*10^7 m/s r = m v / (q B) = (1.67 * 10^-27 kg) *( 2.3 * 10^7 m/s) / ((1.6 * 10^-19 C) * 3.5 T) = .069 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): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qWhat is the radius of orbit for a proton with kinetic energy 5.4 MeV? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Now that the kinetic energy is doubled, the velocity will increase by a factor of sqrt(2) and so will the radius . therefore the radius is sqrt(2) * 0.069 = 0.098 m. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Doubling the KE of the proton increases velocity by factor sqrt(2) and therefore increases the radius of the orbit by the same factor. We end up with a radius of about .096 m. ** STUDENT QUESTION what numbers were used to find this? INSTRUCTOR RESPONSE In Problem 27.60, above, we found the radius of orbit for a proton with kinetic energy 2.7 MeV. Here we are finding the radius for a proton with twice the KE. We could do this in the same manner as before, and we would get the same result. However thinking in terms of the proportionality, as is done here, is both more efficient and more instructive. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qquery univ 27.74 / 27.72 11th edition 28.66 (was 28.52) rail gun bar mass m with current I across rails, magnetic field B perpendicular to loop formed by bars and rails What is the expression for the magnitude of the force on the bar, and what is the direction of the force? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The expression for the magnitude of the force on the bar is F= I*B*L V^2 = V_0^2 + 2ad V^2= 2ad d= v^2/2a a=I*B*L / m d= v^2*m / 2*I*B*L d = (1.12*10^4 m/s)^2 * 25 kg/ (2*2000A *0.5 m* 0.5 T) d= 3.14*10^6 m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Equations needed: f= v/ 2pi*r, I= f*q Now given those equations , we manipulate them both and get I = v*q/ (2pi*r) The area path is pi*r^2 then so magnetic momentum of the quark = I*A= q*v/(2pi*r)*pi*r^2 = q*v*r/2 2/3 e *v*r /(2) + 2 * (1/3 e *v*r /2) = 4/3 e*v*r / 2 = 2/3 e*v*r V = 3/2 mu/ e*r = 3/2 *(9.66*10^-27 Am^2 )/ ((1.6*10^-19 C)*(1.2*10^-15 m))= 7.5 *10^7 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If r is the radius of the orbit and v the velocity then the frequency of an orbit is f = v / (2 pi r). The frequency tells you how many times the charge passes a given point per unit of time. If the charge is q then the current must therefore be I = q f = q v / (2 pi r). Half the magnetic moment is due to the u quark, which carries charge equal and opposite to the combined charge of both d quarks, the other half to the d quarks (which circulate, according to this model, in the opposite direction with the same radius so that the two d quarks contribute current equal to, and of the same sign, as the u quark). The area enclosed by the path is pi r^2, so that the magnetic moment of a quark is I A = q v / (2 pi r) * pi r^2 = q v r / 2. The total magnetic moment is therefore 2/3 e * v r / 2 + 2 ( 1/3 e * v r / 2) = 4/3 e v r / 2 = 2/3 e v r.. Setting this equal to the observed magnetic moment mu we have 2/3 e v r = mu so that v = 3/2 mu / (e r) = 3/2 * 9.66 * 10^-27 A m^2 / (1.6 * 10^-19 C * 1.20 * 10^-15 m) = 7.5 * 10^7 m/s, approx.. Note that units are A m^2 / (C m) = C / s * m^2 / (C m) = m / s. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qquery univ 28.78 / 28.70 11th edition 28.68 (29.56 10th edition) infinite L-shaped conductor toward left and downward. Point a units to right of L along line of current from left. Current I. What is the magnetic field at the specified point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Since one side of the L shaped conductor is parallel to the point, the field of that side will be 0. Therefore, due to the vertical leg with infinite length , the magnetic field at P is B = mu*I/(2pi*a). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: ok