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Information for Physics 1201 Midterm I Wednesday, February 20 My lecture slides are posted at http://www.physics.ohio-state.edu/~humanic/ Information for Physics 1201 Midterm I Wednesday, February 20 1) Format: 10 multiple choice questions (each worth 5 points) and two show-work problems (each worth 25 points), giving 100 points total. 2) Closed book and closed notes. 3) Equations and constants will be provided on the midterm 4) Covers the material in Chapters 16, 17, 18, 19, and 20 Chapter 20 Magnetic Forces and Magnetic Fields Magnetic Fields Produced by Currents A solenoid is made up of many current loops which extend a finite distance along the axis of the loops characterized by the number of turns (number of loops) per unit length, n. The B-field in the interior of a long solenoid depends only on n and the current in the loops, I, as B = µonI Interior of a long solenoid (B ~ 0 outside) B-field lines similar to those of a bar Use RHR magnet with N and S poles shown. for direction of B inside Example: Calculation of the B-field in a solenoid A 0.25 m long solenoid with 5000 turns has a current of 3.5 A applied to it. Find the B-field in the solenoid. B = µonI 5000 turns n = = 2.0 ×104 turns 0.25 m m B = (4π ×10−7 )(2.0 ×104 )(3.5) = 0.088 T = 880 G Magnetic Materials The intrinsic “spin” and orbital motion of electrons gives rise to the magnetic properties of materials è electron spin and orbits act as tiny current loops. In ferromagnetic materials groups of 1016 - 1019 neighboring atoms form magnetic domains where the spins of electrons are naturally aligned with each other; magnetic domain sizes are ~ 0.01 - 0.1 mm. An external magnetic field can induce magnetism in ferromagnetic materials by merging and aligning domains. Depending on the material, the induced magnetism may or may not become permanent. Putting iron in the center of a solenoid can create a strong electromagnet with fields 100x - 1000x the applied fields (also, can turn fields on and off). Ampere’s Law AMPERE’S LAW FOR STATIC MAGNETIC FIELDS For any current geometry that produces a magnetic field that does not change in time, sum up all of the segment lengths Δl multiplied by the B|| è this equals a constant multiplied by the current enclosed by the path. In mathematical notation, B I ∑ ||Δ = µo net current passing through surface bounded by path Note: this is a messy thing to use for anything except in situations where the geometry of the problem has a lot of symmetry, e.g. the long straight wire Ampere’s Law Example: An Infinitely Long, Straight, Current-Carrying Wire Use Ampere’s law to obtain the magnetic field è lots of symmetry! B I ∑ ||Δ = µo Can choose a path where B|| is constant B I (∑ Δ)= µo Since path is a circle of radius r B2π r = µo I On RHS, I is the only current enclosed by path µo I B = è We get our familiar equation! 2π r Example: Ampere’s law applied to a solenoid B I B Δ + B Δ + B Δ + B Δ = µ I ∑ ||Δ = µo ⇒ ( || )ab ( || )bc ( || )cd ( || )da 0 enclosed n turns/length Since, I B ≈ 0 outside solenoid B ⊥ Δ along bc and da d B c then, (B||Δ) = µ0Ienclosed a b cd Use closed path abcda L Since B | | Δ along cd and Ienclosed = nLI as seen earlier ∴BL = µ0nLI ⇒ B = µ0nI The Force on a Current in a Magnetic Field Example: The Force and Acceleration in a Loudspeaker The voice coil of a speaker has a diameter of 0.025 m, contains 55 turns of wire, and is placed in a 0.10-T magnetic field. The current in the voice coil is 2.0 A. (a) Determine the magnetic force that acts on the coil and the cone. (b) The voice coil and cone have a combined mass of 0.020 kg. Find their acceleration. The Force on a Current in a Magnetic Field (a) F = ILB sin θ = (2.0 A)[55π(0.025 m)](0.10 T)sin 90o = 0.86 N F 0.86 N 2 (b) a = = = 43m s m 0.020 kg The Torque on a Current-Carrying Coil Consider the forces on a current-carrying loop in a magnetic field: The two forces on the loop have equal magnitude but an application of RHR shows that they are opposite in direction. The Torque on a Current-Carrying Coil The loop tends to rotate such that its normal becomes aligned with the magnetic field. The Torque on a Current-Carrying Coil Derivation for the net torque on a current-carrying loop in a B field. torque = (force) x (moment arm) A = Lw 1 1 Net torque =τ = ILB(2 wsinφ)+ ILB(2 wsinφ)= IABsinφ L For a coil of N loops (turns) è τ = NIABsinφ = MBsinφ M = NIA Magnetic moment of the coil The Torque on a Current-Carrying Coil Example: The Torque Exerted on a Current-Carrying Coil A coil of wire has an area of 2.0x10-4m2, consists of 100 loops or turns, and contains a current of 0.045 A. The coil is placed in a uniform magnetic field of magnitude 0.15 T. (a) Determine the magnetic moment of the coil. (b) Find the maximum torque that the magnetic field can exert on the coil. (a) M = NIA = (100)(0.045 A)(2.0 ×10−4 m2 ) = 9.0 ×10−4 A⋅ m2 4 2 4 (b) τ = MBsinϕ = (9.0 ×10− A⋅ m )(0.15 T)sin90 =1.4 ×10− N ⋅ m The galvanometer revisited How a galvanometer works. The coil of wire and pointer rotate when there is a current in the wire. The pointer stops when the magnetic torque on the coil, τ, is canceled by the opposing torque of the spring, τs . τ M = NIABsinφ τ = kΔφ Δφ is the angular displacement of the needle S k is the spring constant (Hooke’s Law) ∑τ = τ M −τ S = 0 NIABsinφ NIABsinφ − kΔφ = 0 ⇒ Δφ = ⇒ Δφ ∝ I k The Torque on a Current-Carrying Coil The basic components of a dc motor. Split-ring commutator } The Torque on a Current-Carrying Coil How a dc motor works. When a current exists in the coil, Because of its inertia, the coil the coil experiences a torque. continues to rotate when there is no current. .
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