PHYS 012 1st Edition Lecture 12Outline of Last Lecture I. Magnetic Fieldsa. Magnets have two poles: north and south.b. Like poles repel, opposites attract.c. No magnetic monopoles have been found.d. Magnetic fields exist near moving charge.e. Field lines come out of north and go to south poles.f. Magnetic field at a point is tangent to field line (B).II. Earth’s Magnetic Fielda. Geographic north is magnetic south (but this switches every couple ten thousandyears).b. Compass will point in direction of magnetic field lines (ie. geographic north).c. Magnetic north is a few degrees off of “true north.”III. Magnetic Forcea. Electric charges moving in a magnetic field can feel a force.i. F = qvB sinθ1. v = velocity2. B = magnetic field strength in teslas (T) = (N*s)/(C*m)3. θ = angle between v and B if they are tail to tailb. Force is always perpendicular to v and B.c. Movement of a particle in magnetic field will be circular.i. ΣF = qvBsinθ = macii. sin (90°) = 1iii. ac = v2/riv. qvB = macv. qvB = mv2/rvi. r = mv/qBd. Right-hand Rule:i. 1. Hold right hand so fingers point in the direction of v (opposite directionif charge is negative)ii. 2. Bend fingers in direction of B.iii. 3. Thumb will be pointing in direction of F.iv. vector going toward youv. ✕ vector going away from youIV. Magnetic Force versus Electric Forcea. FE acts along direction of E fieldb. FB acts perpendicular to B fieldThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.c. FE acts on stationary or moving chargesd. FB acts on moving charges onlye. FE does work in displacing a charged particlef. FB from steady doesn’t do worki. W = Fscosθii. F always perpendicular to viii. cos (90°) = 0iv. W = Fs (0) = 0Outline of Current Lecture V. Velocity Selectora. The electric field between a parallel plate capacitor is pointing downward with a force equal to qE.b. The magnetic field coming toward observer with a force equal to qvB (pointing upward).c. The particle will move straight through if the forces are equal.i. qvB = qEii. v = E/B (faster = curves upward, slower = curves downward)VI. Problem: A particle accelerates from rest through a potential difference and then enters a magnetic field. Find an expression for the mass of the particle.a. ΔEPE = ΔKEb. qΔV = ½ mv2c. v = [(2q|ΔV|)/m]½ d. ΣF = qvB = mace. qB = m/r [(2q|ΔV|)/m]½f. m = (r2B2q)/(2ΔV)VII. Current through a straight conducting wire in a uniform magnetic fielda. Electrons moving downward, feel force to the lefb. F = ILBsinθVIII. Torque on a rectangular electrical circuita. F1 = -F3 = 0b. F2 = -F4 Side view of circuitc. τ = rFsinθd. τnet = τ2 + τ4 = (b/2)(IaB)(sin(90°)) + (b/2)(IaB)(sin(90°)) = abIB = AIBi. A = area of circuit = abe. Generally, τnet = IABsinθf. With multiple loops, τnet = NIABsinθi. N = number of
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