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screen menu, calcs & info menu buttons # shows 0 volts for each channel cursor control autoset measure cursor triggering control usual channel controls Ideal Solenoid B Boutside=0 B B B uniform magnetic field inside Boutside=0 Use Ampere’s Law to Find Magnetic Field (Explain each step in your report.) B ds I o total enclosed L Amperian Loop whole loop B ds NI o BIN inside solenoid L 0 Bin dy o NI Bin 0 dy o NI L Bin L o NI Bin N solenoid loops enclosed, each with current I. o NI L Bin o nI where n is “loop density” N/L of solenoid. Cause and Effect in a Solenoid: Ampere’s Law to Faraday’s Law voltage VR=RIR=RIL The current in the solenoid creates a magnetic field inside the solenoid due to Ampere’s Law. Bampere t The changing magnetic field inside the solenoid L causes a back EMF (voltage) due to Faraday’s Law. Notice that dI/dt causes a phase shift. Inside the solenoid: dB 0 dt B S N V(velocity) R Direction of current inside the resistor? Transmitter Oscillating transmitting magnetic fields. Oscillating transmitting voltage. Receiver Transmitting magnetic fields reach inside coils. Oscillating voltage received is measurable. The LRC Circuit - AC Driven voltage VR VL VC The LRC Circuit - AC Driven: Source from Addition Vsource voltage VR VL VC A B C 100 50 mH 0.1 F Iamplitude Iamplitude Large R Small R fdrive fresonance fdrive fresonance VR(t) VR(t) VS(t) out of phase 45o in phase VS(t) System: Charged hollow sphere with inner radius a and outer radius b. Charges: nonuniform charge distribution in between (so not a conductor): A (r) r Problem: The electric field is a radial vector field due to the symmetry of the system. Find the electric field magnitude in the radial direction at every distance from the origin. Required vector calculus knowledge: dQ dV charge volume rI r2 2 0 0 d sin d r3 A 4 r 2 d r 4 Ardr Problem solving strategy: 1) Draw non-physical Gaussian sphere at distance r where you want to find Er. 2) Use Gauss’s law to write equation for Er in terms of other parameters. 3) Solve for Er. In this case solve in 3 places, inside hollow region (rI), inside charged region (r2) and outside (r3). System: Charged infinite cylinder with radius a. Charges: Nonuniform charge distribution inside cylinder (so not a conductor): Problem: The electric field is a radial vector field due to the symmetry of the system. Find the electric field magnitude in the radial direction at every distance from the origin. Required vector calculus knowledge: 2 0 dQ dV d dz rdr (r) Ar charge volume Try solving over a finite height zo: dQ dV charge volume 2 zo 0 0 d dz rdr 2zo Arrdr rI r2 2zo Ar 2 dr Problem solving strategy: 1) Draw non-physical Gaussian cylinder at distance r where you want to find Er. 2) Use Gauss’s law to write equation for Er in terms of other parameters including an arbitrary height zo. 3) Solve for Er. In this case solve in 2 places, inside region (rI), and outside (r2). You will need to have the arbitrary height zo cancel in the end. Another view of drawing a Gaussian cylinder of radius r and finite length zo around an infinite cylinder of charge (this one outside). zo r System: Charged infinite slab of width w in x-y direction. Required vector calculus knowledge: Charges: Uniform slab of charge density : dQ dV dx dy dz charge Problem: The electric field is a vector field pointing perpendicular to the plane of the slab due to the symmetry of the system. Find the electric field magnitude in the perpendicular direction at a given distance from the middle of the slab. xo z1 z2 dQ dV charge volume xo yo z 0 0 0 dx dy d z x o y o dz yo Try solving over a finite box xo and yo: xo yo volume 0 Set z=0 in middle of sla Problem solving strategy: 1) Draw non-physical Gaussian rectangular prism from center of slab to height z where you want to find Ez. 2) Use Gauss’s law to write equation for Ez in terms of other parameters including arbitrary length and width xo and yo. 3) Solve for Er. In this case solve in 2 places, inside region (zI), and outside (z2). You will need to have the arbitrary xo and yo cancel in Three representations of the same circuit: V V + BATTERY V BATTERY 0 BATTERY (Note: bulb shape distorted.) 3.0 + + Circuit Position bulb shape distorted.) d c BATTERY d d c BATTERY c ATTERY c b a BATTERY 3V + b a + 3V d + A. B. C. 1.5 V 1.5 V 1.5 V 1.5 V .5 V A. B. C. 1.5 V .5 V 1.5 V 1.5 V 1.5 V A. .5 V B. + V - 1.5 V D C. + V - 1.5 V 1.5 V + 1.5 V V - ITOTAL BATTERY + ID BATTERY BATTERY IE IC IB IA + + + BATTERY BATTERY BATTERY + BATTERY 3V + + Measuring the voltage drop across a light bulb (DMM in parallel) Voltage VDC R V Measuring the voltage drop across a light bulb (DMM in series): Amperes mA R A Measuring the resistance of a light bulb (component disconnected): Ohms () R V 1.5 + BATTERY 0 Circuit Position V 1.5 + BATTERY Circuit Position Right-hand-wrap rule for finding direction of magnetic poles created by moving charges (current) Thumb points to North N q f charge is negative, everse poles. Wrap fingers in direction of current. QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. N S N S N S S S SMAGNETIC N S NMAGNETIC Excess positive charge on the surface of a sphere. + + - -- - + + + + + + + + + Ex.1. Excess negative charge on the surface of a cube. + + + Ex.2. - - - - - (None of the excess charges rest inside the objects, they always repel each other to the surface.) Macroscopic charge separation across a neutral conductor in the presence of an electric field. - + - + ++ + + + + - - -- - - E Negatively charged object creates an electric field. - Microscopic charge separation across a neutral conductor in the presence of an electric field. Positively charged object creates an electric field at surface of material. + E + + rotate + insulator/dielectric material Electrons deposited on the surface of a balloon by rubbing it against your hair do not spread out. - -- -