course phy 122 ïOÀï€r¢g¸ûîÍÌð—Ｊz³ý¢ŠÂassignment #009
......!!!!!!!!...................................
18:57:30 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.
......!!!!!!!!...................................
RESPONSE --> the force is related to the velocity, in that the feild can either accelerate, decelerate or hold steady a particle. he feild will exert force on the particle, and as the particle has to express the energy exerted upon it, it can either chane chemically, thermally, or motivally. most usually, it moves, and so the velocity is related to the feild
.................................................
......!!!!!!!!...................................
18:58:49 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ah, I apporached the problem form the viewpoint of does the feild affect velocity, rather than does the velocity have in imapct on the feild
.................................................
......!!!!!!!!...................................
18:59:40 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.
......!!!!!!!!...................................
RESPONSE --> the force on a charged particle due to a magnetic feild is not affected by it velocity, rather, it is affected by it's position relative to the field
.................................................
......!!!!!!!!...................................
19:00:05 ** 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. **
......!!!!!!!!...................................
RESPONSE --> k
.................................................
......!!!!!!!!...................................
19:03:48 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.
......!!!!!!!!...................................
RESPONSE --> velocity is a function of mass and force, with accleration thrown in for good measure, and as such, the force applied does have an impact on velocity. as to the net force, the charged particle has a constant force bewing played upon it, and so the velocitty tends to increase or decrease, due to the force pulling/pushing it. (think gravity, it doesn't change, but a dropped object's V is higher the longer it falls)
.................................................
......!!!!!!!!...................................
19:04:06 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ah
.................................................
......!!!!!!!!...................................
19:06:49 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.
......!!!!!!!!...................................
RESPONSE --> force is I * L * b I*L is 24000 tessla, but it is at an angle, with another force, B 24000t * b is 1.32 times sin (angle), 1.32sin(65)= 1.20
.................................................
......!!!!!!!!...................................
19:06:56 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).
......!!!!!!!!...................................
RESPONSE --> cool
.................................................
......!!!!!!!!...................................
19:13:14 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.
......!!!!!!!!...................................
RESPONSE --> its gonna spin. this can be found by force is qvb, which is 1.6*10^-19C * 8.75*10^5* .75 this works out to give force of 1.05E-13 but you still need ncentripetal acceleration, which comes to centripital force/ mass 1.05*10^-13/9.11*10^-31= 1.15E-45
.................................................
......!!!!!!!!...................................
19:13:29 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:14:12 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?
......!!!!!!!!...................................
RESPONSE --> I will solve
.................................................
......!!!!!!!!...................................
19:14:40 **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. **
......!!!!!!!!...................................
RESPONSE --> ok, noted
.................................................
......!!!!!!!!...................................
19:14:45 query univ 27.60 (28.46 10th edition). cyclotron 3.5 T field.
......!!!!!!!!...................................
RESPONSE --> principles
.................................................
......!!!!!!!!...................................
19:14:56 What is the radius of orbit for a proton with kinetic energy 2.7 MeV?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:15:01 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:15:06 What is the radius of orbit for a proton with kinetic energy 5.4 MeV?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:15:08 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:15:11 query univ 28.73 (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?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:15:15 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:15:19 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?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:15:22 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
"