Physics 2102 Jonathan Dowling James Clerk Maxwell 1831 1879 Lecture 34 MON 13 APR Ch 33 1 3 5 7 E M Waves QuickTime and a decompressor are needed to see this picture MT03 Avg 65 100 P3 K Schafer Office hours MW 1 30 2 30 pm 222B Nicholson Q2 J Dowling Office hours MWF 10 30 11 30 am 453 Nicholso Q3 M Gaarde Office hours TTh 2 30 3 30 pm 215B Nicholson Q2 C Buth Office hours MF 2 30 3 30 pm 222A Nicholson A 90 100 B 80 89 C 60 79 D 50 59 Problem 1 18 points In the figure below two semicircular arcs I II have radii R1 4 10 cm and R2 6 60 cm carry current i 0 281 A and share the same center of curvature C The same current i also flows through the straight sections of wire labeled III IV Question 2 8 points The figure below shows four arrangements in which long parallel wires carry equal currents i directly into or out of the page at the corners of identical squares a 5 pts What is the contribution to the magnitude of the magnetic field at point C from the two straight sections of wire III and IV Explain your answer BIII BIV 0 r r r dB ds r 0 Biot Savart b 8 pts Calculate the magnitude of the magnetic field at the point C due to all four sections of wire Out r r i 1 1 BTOT BI BII 0 8 14 10 7 T 4 RI RII half a circle Left c 5 pts What is the direction of the total magnetic field at the point C due to all four sections of wire Circle one In Up Right Down Fieldis Zero Right Hand Rule Vector Addition Right Hand Rule BI BII Since I is Closer Out of the page B r 0 i 4 pts For square B what is the direction of the magnetic field with respect to the page at the center of the square Circle one Into the page BA BB BC BC BB BA BB BC BA BA BB BC BA BC BB Up towardsthe topof the page Down towardsthe bottom of the page To the right of the page To the left of the page The total magnetic feld at C is zero and hasno direction ii 4 pts Rank the sections according to the magnitudeof the magnitudeof the magnetic feld at the centerof each square grea test frst Circle one B 0i 2 r BA 0 BB 2 2B BC 2B Maxwell Waves and Light A solution to the Maxwell equations in empty space is a traveling wave d B ds 0 0 E dA dt S C d E ds B dA dt S C ectric and magnetic forces can travel d2E d2E 0 0 2 E E0 sink x ct 2 dx dt 1 c 3 108 m s 0 0 The electric waves trav at the speed of light Light itself is a wave o electricity and magnetis Electromagnetic Waves A solution to Maxwell s equations in free space E Em sin k x t B Bm sin k x t c speed of propagation k Em 1 c Bm 0 0 m 299 462 954 187 163 miles sec s Visible light infrared ultraviolet radio waves X rays Gamma rays are all electromagnetic waves QuickTime and a decompressor are needed to see this picture Radio waves are reflected by the layer of the Earth s atmosphere called the ionosphere This allows for transmission between two points which are far from each other on the globe despite the curvature of the earth Marconi s experiment discovered the Maxwell s Rainbow The wavelength frequency range in which electromagnetic EM waves light are visible is only a tiny fraction of the entire electromagnetic spectrum Fig 33 2 Fig 33 1 33 The Traveling Electromagnetic EM Wave Qualitatively An LC oscillator causes currents to flow sinusoidally which in turn produces oscillating electric and magnetic fields which then propagate through space as EM waves Next slide Fig 33 3 Oscillation Frequency 1 LC 333 Mathematical Description of Traveling EM Waves E Electric Field Em sin kx t B Magnetic Field 1 Wave Speed c 0 0 Bm sin kx t All EM waves travel a c in vacuu 2 Wavenumber k c EM Wave Simulation 2 2 f Frequency T 0 Vacuum Permittivity 0 Vacuum Permeability Fig 33 5 E Amplitude Ratio m c Bm E t c Magnitude Ratio B t 33 The Poynting Vector Points in Direction of Power Flow omagnetic waves are able to transport energy from transmitte eiver example from the Sun to our skin ower transported by the wave and its tion is quantified by the Poynting vector John Henry Poynting r 1 r r S E B 0 For a wave since E is perpendicular to B S 1 1 2 EB E 0 c 0 In a wave the fields change with time Therefore the Poynting vector changes too Units Watt m2 E S B 1852 1914 The direction is constant but the magnitude changes from 0 to a maximum EM Wave Intensity Energy Density A better measure of the amount of energy in an EM wave is obtained by averaging the Poynting vector over one wave cycle The resulting quantity is called intensity Units are 2 also Watts m 1 2 1 2 2 1 The average of sin2 over I S Emsin kx t Em E c 0 c 0 2c 0 one cycle is Em2 1 I Em 2 or 2c 0 Both fields have the same energy density 1 I Erms 2 c 0 1 1 1 B2 1 B2 2 2 uE 0E 0 cB 0 uB 2 2 2 0 0 2 0 e total EM energy density is then 2 2 u 0 E B 0 2 Erms Solar Energy The light from the sun has an intensity of about 1kW m2 What would be the total power incident on a roof of dimensions 8m x 20m 1kW m2 is power per unit area IA 103 W m2 x 8m x 20m 0 16 MegaWatt The solar panel shown BP 275 has dimensions 47in x 29in The incident power is then 880 W The actual solar panel delivers 75W 4 45A at 17V less than 10 The electric meter on a efficiency solar home runs backwards QuickTime and a decompressor are needed to see this picture EM Spherical Waves The intensity of a wave is power per unit area If one has a source that emits isotropically equally in all directions the power emitted by the source pierces a larger and larger sphere as the wave travels outwards 1 r2 Law Ps I 2 4 r So the power per unit area decreases as the inverse of distance Example A radio station transmits a 10 kW signal at a frequency of 100 MHz At a distance of 1km from the antenna find the amplitude of the electric and magnetic field strengths and the energy incident normally on a square plate …
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