PHY2049_06-15-11

# PHY2049_06-15-11 - Chapter 31 A horizontal power line...

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6/15/2011 1 Chapter 31 Electromagnetic Oscillations and Alternating Current In this chapter we will cover the following topics: -Electromagnetic oscillations in an LC circuit -Alternating current (AC) circuits with capacitors -Resonance in RCL circuits -Power in AC-circuits -Transformers, AC power transmission (31 - 1) A horizontal power line carries a current of 5000 A from south to north. Earth's magnetic eld (60 μ T) is directed towards north and inclined down-ward 50 degrees to the horizontal. Find the magnitude and direction of the magnetic force on 100 m of the line due to the Earth's field. (1)23 N, west (2) 23 N, east (3) 30N, west (4) 30N, east (5) none of these 50 B N I Suppose this page is perpendicular to a uniform magnetic field and the magnetic flux through it is 5Wb. If the page is turned by 30 around an edge the flux through it will be: A. 2.5Wb B. 4.3Wb C. 5Wb D. 5.8Wb E. 10Wb A car travels northward at 75 km/h along a straight road in a region where Earth’s magnetic field has a vertical component of 0.50 × 10 -4 T. The emf induced between the left and right side, separated by 1.7m, is: A. 0 B. 1.8mV C. 3.6mV D. 6.4mV E. 13mV L C The circuit shown in the figure consists of a capacitor and an inductor . We give the capacitor an initial chanrge and then abserve what happens. The capacitor will discharge th C L Q LC Oscillations rough the inductor resulting in a time dependent current . i We will show that the charge on the capacitor plates as well as the current 1 in the inductor oscillate with constant amplitude at an angular frequency The total energy in the circuit is t q i LC U ϖ = 2 2 he sum of the energy stored in the electric field of the capacitor and the magnetic field of the inductor. . 2 2 The total energy of the circuit does not change with time. Thus E B q Li U U U C dU = + = + 2 2 2 2 0 0. 1 0 dt dU q dq di dq di d q Li i dt C d d q L t dt dt q dt dt dt C + = = + = = = = (31 - 2) L C 2 2 2 2 1 0 ( ) This is a homogeneous, second order, linear differential equation which we have encountered previously. We used it to d 1 0 escribe the simple harmonic oscillat o d q L q dt C d q q dt LC + = = + eqs.1 2 2 2 0 with sol r (SHO) ( ) ution: ( ) cos( ) d x x dt x t X t ϖ ϖ φ + = = + eqs.2 ( 29 If we compare eqs.1 with eqs.2 we find that the solution to the differential equation that describes the LC-circuit (eqs.1) is: 1 ( ) cos where , and is the phase angle. The current q t Q t LC ϖ φ ϖ φ = + = ( 29 sin dq i Q t dt ϖ ϖ φ = = - + ( 29 ( ) co s q t Q t ϖ φ = + 1 LC ϖ = (31 - 3)

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6/15/2011 2 L C ( 29 ( 29 ( 29 2 2 2 2 2 2 2 2 2 2 The energy stored in the electric field of the capacitor cos 2 2 The energy stored in the magnetic field of the inductor sin sin 2 2 2 The total energy 2 E B E B q Q U t C C Li L Q Q U t t C U U U Q U ϖ φ ϖ ϖ φ ϖ φ = = + = = + = + = + = ( 29 ( 29 2 2 2 cos sin 2 The total energy is constant; Q t t C C ϖ φ ϖ φ + + + =   energy is conserved 2 2 3 The energy of the has a value of at 0, , , ,...
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