MIT Department of Physics8.02 Spring 2014Problem Set 8Due: Tuesday, April 15 at 9 pm.Submit the two online problems in this set (problems 1 and 2) online. Hand in the six writtenproblems in this set (problems 3-8) in your section slot in the boxes outside the door of 32-082 or26-152 depending on which is your classroom. Make sure you clearly write your name and sectionon your problem set.Text: Dourmashkin, Belcher, and Liao; Introduction to Electricity and Magnetism MIT 8.02 CourseNotes Revised.Week Ten: Magnetic Inductionand Energy; DC CircuitsProblem Set 7 Due Tuesday April 8 at 9 pmW10D1 M/T Apr 7/8ReadingInductance & Magnetic EnergyCourse Notes: Sections 11.1-11.3W10D2 W/R Apr 9/10ReadingDC Circuits & Kirchhoff's Loop RulesCourse Notes: Sections 7.1-7.5W10D3 F Apr 11ReadingProblem Solving 7: Building a CircuitCourse Notes: Sections 7.1-7.5, 7.10Week Eleven: RC, LR, andUndriven RLC CircuitsProblem Set 8 Due Tuesday April 15 at 9 pmW11D1 M/T Apr 14/15ReadingRC CircuitsCourse Notes: Sections 7.7-7.8, 7.11.3W11D2 W/R Apr 16/17ReadingRL Circuits; Experiment 4: Part 1: RC and LR CircuitsCourse Notes: Sections 11.4-11.6, 11.12.2, 11.13.4-11.13.5W11D3 F Apr 18ReadingProblem Solving 8: RC and RL CircuitsCourse Notes: Sections 7.11.3, 11.12-11.13.5Problem 1: Current, Energy and PowerThis problem must be submitted online.A battery of emf has internal resistance . Suppose that the battery is connected by wires withnegligible resistance to an external (load) with resistance .Note that the quantity of charge delivered by the battery is a given constant and you may use it inyour answer.(Part a) What value of maximizes the total power delivered to the load?(Part b) How much energy is converted in the load before it expires?Load Energy =(Part c) How much energy is lost to heat in the internal resistance of the battery?Energy Lost in Battery=RiRLQRL=RLProblem 2: Superconducting MagnetsThis problem must be submitted online.To make a strong, uniform magnetic field, a superconducting solenoid is frequently employed. Oncecurrent is “loaded" into the solenoid, the two ends can be tied together (we'll look at how this is donein next week's problem set) and then the current will continue to circulate forever. Calculate theinductance and total energy stored in the following two magnets. Ignore the fields outside themagnets.(Part a) The strongest constant field laboratory magnets currently available create a field of 45 Twith a current of 150 A. They have a bore diameter of 5 cm and an active length (where the field isbasically uniform) of about 10 cm.(in )(in )(Part b) An MRI magnet is 2 m long, 0.75 m in diameter and has a field strength of 4 T with acurrent of 100 A. In designing MRI magnets, people work very hard to make the field uniformthroughout the core of the magnet (that is to say, it's not just a straight solenoid).(in )(in )L =HU =JL =HU =JProblem 3: Self Inductance, Energy, Induced Electric Fields,and Faraday's LawYou must hand in a written solution to this problem. You may check your answers online ifyou wish, but this is not required.A wire is wrapped times around a cylinder of non-magnetized material of radius cm and length 25 cm. At , the current through the wire is increased according towhere A/s. . After s the current remains constant. A small wire loop of radius cm is placed on the solenoid's axis at the mid point of the solenoid with the normal to the plane ofthe loop pointing along the solenoid's axis. The loop has resistance .(Part a) After s , use Ampere's Law to find the direction and magnitude of the magnetic fieldwithin the solenoid. (Ignore edge effects).(in T)(Part b) Sketch the magnetic field lines everywhere including edge effects.(Part c) What is the self-inductance of the solenoid?(in H)(Part d) The rate of doing work against the back emf is , where the emf is given by . How much work is done against the back emf in order to reach a steady current after s in the solenoid?(Part e) The stored energy in a magnetic field is equal to . How muchenergy is stored in the magnetic field of the solenoid? Do you expect your result to agree or disagreewith your answer in part d)? Briefly explain your reasoning.(Part f) The changing magnetic flux inside the solenoid produces an induced tangential electricfield. During the interval, s , find an expression for the induced electric field as a functionof distance from the central axis of the solenoid .N = 100a = 3.0l =t = 0I(t) = bt, 0 < t < 2sb = 0.2t > 2r = 1.0R = 5.0Ωt > 2B =L == −IdWdt = −LdIdtt > 2W == dVUmagnetic12μ0∫all spaceB2=Umagnetic0 < t < 2r| (r)| =EindProblem 4: Mutual InductanceYou must hand in a written solution to this problem. You may check your answers online ifyou wish, but this is not required.The figure shows a solenoid of radius and length located inside a longer solenoid of radius and length . The total number of turns is on the inner coil, on the outer.(Part a) What is the mutual inductance ?(Part b) If a current flows through the little solenoid, what is the total flux through the big one?a1b1a2b2N1N2MM =I1=Φ21Problem 5: Self-Inductance of Two WiresYou must hand in a written solution to this problem. You may check your answers online ifyou wish, but this is not required.Two long parallel wires, each of radius , whose centers are a distance apart carry equal currentsin opposite directions. What is the self-inductance of a length of such a pair of wires? You mayneglect the flux within the wires themselves.adlL =Problem 6: GeneratorYou must hand in a written solution to this problem. You may check some of your answersonline if you wish, but this is not required.A "pie-shaped" circuit is made from a straight vertical conducting rod of length welded to aconducting rod bent into the shape of a semi-circle with radius (see sketch). The circuit iscompleted by a conducting rod of length pivoted at the center of the semi-circle, (at point ), andfree to rotate about that point. This moving rod makes electrical contact with the vertical rod at oneend and the semi-circular rod at the other end. The angle is the angle between the vertical rod andthe moving rod, as shown. The circuit sits in a constant magnetic field pointing out of the page.(Part a) If the angle is increasing with time, what is the direction of the resultant current flowaround the "pie-shaped" circuit?For the next two parts,
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