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Course: PHYS 251, Fall 2008
School: UVA
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251/Fall PHYS 1998 Homework #8 Solutions Due: Friday October 30, 1998 Page 8-1 1 (Tipler 26-47). (a) The flux through the loop is = B (effective area) = B N ab cos , where = t is the angle between the loop normal and the magnetic field B. From Faraday's law, E =- d d = - (N Bab cos t) = N Bab sin t. dt dt (1) Note that the problem does not give you an explicit for for , which could just as easily have been...

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251/Fall PHYS 1998 Homework #8 Solutions Due: Friday October 30, 1998 Page 8-1 1 (Tipler 26-47). (a) The flux through the loop is = B (effective area) = B N ab cos , where = t is the angle between the loop normal and the magnetic field B. From Faraday's law, E =- d d = - (N Bab cos t) = N Bab sin t. dt dt (1) Note that the problem does not give you an explicit for for , which could just as easily have been defined as t - , with as any angle you wish to choose. Hence the "sin" in the derived equation could just as easily been " sin" or " cos". (b) The amplitude of the emf out of this generator is E0 = N Bab. To have E0 = 110 V, with the given values for the other parameters, we must have an angular frequency = E0 /N Bab = (110)/[(1000)(2)(0.01)(0.02)] = 275 rad/s. 2 (Tipler 26-58). There are a large number of ways to solve this problem. One (probably) very bad way is to look at the total magnetic flux 1 + 2 . Unfortunately, since the book mentions flux explicitly, you may have been given the impression that this quantity is in some way a key to the solution you seek. The statement ". . . none of the flux from either passes through the other..." is meant to signify that the flux through inductor one (for example) is not given by 1 = L1 I1 + M I2 , (2) which would mean that there is some mutual inductance between the two components. So, because we may write 1 = L1 I1 and 2 = L2 I2 , we can also write that the voltage drop across inductor one is V1 = L1 (dI1 /dt), and similarly for the second inductor. As it so happens, when the inductors are in parallel, their voltage drops are the same (Kirchoff's first law): V = L1 dI1 dI2 = L2 . dt dt (3) We would define an effective inductance Leff using the relation Leff = V dI dt -1 , or 1 1 dI , = Leff V dt (4) where I = I1 + I2 is the total current that goes into our two-inductor system. From Kirchoff's second law, I = I1 + I2 , and so, dI1 dI2 dI = + . (5) dt dt dt We can use eq. (3) to substitute for the dIk /dt terms: 1 1 = Leff V 3 (Tipler 26-67). dI2 dI1 + dt dt = 1 V V V + L1 L2 = 1 1 + . L1 L2 (6) x E (a) The crossbar has a current that goes down the page, when the battery is attached as described in the problem. A current I flows through the bar, and therefore it will feel a force F = I B that is to the right, PHYS 251/Fall 1998 Page 8-2 for a magnetic field B into the page. The problem now is, of course, to find the current that flows. Normally (no magnetic field), the current would simply be E/R. The problem is that since the circuit is physically expanding, it's magnetic flux is changing and therefore, by Faraday's law, an emf is induced. The size of this "back"-emf, called "back" because it will always oppose the emf applied, is given by |Eback | = d dx d = (B x) = B = B v, dt dt dt (7) where x is the displacement of the crossbar bar (the placement of the origin turns out to not be very important), and v = dx/dt is the velocity. The total emf is therefore E - |Eback | = E - B v, and the current that flows is I = (E - B v)/R. The force on the rod is therefore F =I B= (E - B v) B , R (8) and Newton's second law, F = ma = m(dv/dt), is written m (E - B v) B dv = . dt R (9) (b) The terminal velocity vt is defined the as point at which (dv/dt)v=vt = 0. From above, 0= (E - B vt ) B R vt = E . B (10) (c) The current when v = vt is given by I = (E - B vt )/R = 0. This can be done without resort to algebra if you remember that F = I B and that the force F = 0 because it reached terminal velocity, so we must have I = 0. Now is a good time to look back at a similar problem you had in chapter 24 and ask yourself how this problem is different. 4 (Tipler 26-80). (a) The setup of this problem is very similar to that of Example 25-10 in Tipler. You should know that the symmetry of the problem is such that the magnitude of the B field is constant along circles that are concentric with the cable and have surface normals along the axis of the cable. Three examples of such curves are drawn (dotted) in the accompanying figure. In addition, the direction of the B field at all points along that circle is tangent to it, and has a sense given by the right hand rule. We therefore have that B d = B d at all points along the curve. It is also for that reason that we choose our curves to travel PHYS 251/Fall 1998 Page 8-3 clockwise (as drawn), since the current inside is into the page. With all of these preliminaries out of the way, we can make the seemingly simple statement that one side of Amp`re's law is given by e B d = B 2r. (11) C Now the right hand side of Amp`re's law is multivalued: e 0 I 0 IC = 0 I + (-I) = 0 r < r1 r1 < r < r2 . r > r2 (12) Note that in the last line above, IC has two parts because for r > r2 , C encloses both currents. The inner current goes into the page, which is the same direction as the surface normal of C (which is given by the right hand rule), so it counts as positive. The outer current is coming out of the page, is anti-parallel to the surface normal of C, and therefore counts as negative. Putting all of Amp`re's law together, we get e r < r1 0 0 I (13) B= r1 < r < r2 . 2r 0 r > r2 (b) The magn...

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