course Phys 202 ĕ㷚ݴ}GιǤassignment #030
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15:39:15 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 --> The force is proportional to the velocity. If the velosity is increases,the force increases.
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15:39:56 ** 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 --> I was completely off. I wasn't thinking about a capacitor not creating a magnetic field.
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15:44:08 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 --> It is affected because the force is proportianl to the velocity.
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15:45:17 ** 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 --> I understand that and the equation backs up what I was trying to say.
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16:46:57 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 --> The velocity (v=E/B) is independant of the electric filed force (f=q*E). The forces present are the Electric field, and the magnetic field. The Magnetic field (F=q*v*B) does depend on the velocity of the particle. The magnetic field produces a force is perpindicular to the velocity and the electric field's force. Therefore the forces are equal to each other and cancel each other out.
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16:47:55 ** 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 --> I deduced the same answer and found the net force to be zero also.
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17:59:04 **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. **
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RESPONSE --> Using the centripical force equation (m* (v^2/r)) we should be able to find the velocity. It was here that I got stuck. Because of the gravitational simularities, we have to use coulomb's Law. m*v^2/r=k*(Q1*Q2)/r^2 and solve for v. v=sqrt(k*(Q1*Q2)/(r*m)) v=sqrt(8.98*10^9*(1.6*10^-19*1.6*10^-19)(.529*10^-10*9.11*10^-31)=2.19*10^6m/s The current then can be found using I=dq/dt. The time to orbit is the time it takes to go around the orbit once which is 2*pi*r. v=distance/time dt=(2*pi*.529*10^-10)/2.19*10^6=1.52*10^-16sec I=dq/dt=1.602*10^-19/1.52*10^-16=.001054 amp B=k*2*pi*I/r=2*pi*.001054amps/.529*10^-10=12.5
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17:59:07 query univ 27.60 (28.46 10th edition). cyclotron 3.5 T field.
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18:00:46 query 20.73 (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?
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RESPONSE -->
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18:00:50 ** 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. **
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18:00:57 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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18:00:59 ** 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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