PHY 2054 Fall 2012 Kumar/Mitselmakher Exam I
... top vertex contains a charge (8, 32) nC. The two vertices at the base contain a charge of -4 nC.. Calculate tthe he magnitude (in N/C) and direction of the electric field at point P, the midpoint of the base. Answer (4.5,18), down Solution: Note here that the electric field due to the lower two nega ...
... top vertex contains a charge (8, 32) nC. The two vertices at the base contain a charge of -4 nC.. Calculate tthe he magnitude (in N/C) and direction of the electric field at point P, the midpoint of the base. Answer (4.5,18), down Solution: Note here that the electric field due to the lower two nega ...
Paper
... they can share one beam). This implies that the synthetic magnetic field can now be chosen to be the same, to be opposite, or to be different for the two spin states. One option is to have zero synthetic magnetic field for one of the states. Atoms in this state can still tunnel along the tilt direct ...
... they can share one beam). This implies that the synthetic magnetic field can now be chosen to be the same, to be opposite, or to be different for the two spin states. One option is to have zero synthetic magnetic field for one of the states. Atoms in this state can still tunnel along the tilt direct ...
Bose–Einstein condensation NEW PROBLEMS
... A. De Broglie wavelength and BEC In the semiclassical approximation at low density and high temperature, atoms are localized to wave packets with dimensions small compared to the average interatomic separation. The average de Broglie wavelength l 5 h/p, which is a quantum measure of delocalization o ...
... A. De Broglie wavelength and BEC In the semiclassical approximation at low density and high temperature, atoms are localized to wave packets with dimensions small compared to the average interatomic separation. The average de Broglie wavelength l 5 h/p, which is a quantum measure of delocalization o ...
Physics Week 5(Sem. 2) Magnetism
... moving with a velocity of v through a magnetic field. The vector of the magnetic field is labeled B, it is assumed to be constant in magnitude and direction. If the charge moves parallel or antiparallel then the charge experiences no magnetic field. If, the charge moves perpendicular to the ...
... moving with a velocity of v through a magnetic field. The vector of the magnetic field is labeled B, it is assumed to be constant in magnitude and direction. If the charge moves parallel or antiparallel then the charge experiences no magnetic field. If, the charge moves perpendicular to the ...
Chapter 23 Electrical Potential
... metal sphere that has a charge –Q. Sketch the electric field lines and equipotential surfaces for this system of charges. Picture the Problem The electric field lines, shown as solid lines, and the equipotential surfaces (intersecting the plane of the paper), shown as dashed lines, are sketched in t ...
... metal sphere that has a charge –Q. Sketch the electric field lines and equipotential surfaces for this system of charges. Picture the Problem The electric field lines, shown as solid lines, and the equipotential surfaces (intersecting the plane of the paper), shown as dashed lines, are sketched in t ...
Use of Superconductors in the Excitation System of
... In the mixed state, the magnetic field is able to penetrate inside the bulk of the superconductor. This penetration only exists in tiny tubes known as flux tubes (Figure II-4 (a)), which are surrounded by superconducting current vortices, as shown in Figure II-4 (b). The flux tubes are enclosed in a ...
... In the mixed state, the magnetic field is able to penetrate inside the bulk of the superconductor. This penetration only exists in tiny tubes known as flux tubes (Figure II-4 (a)), which are surrounded by superconducting current vortices, as shown in Figure II-4 (b). The flux tubes are enclosed in a ...
Adobe Acrobat file () - Wayne State University Physics and
... By a clever change to the rings and brushes of the ac generator, we can create a dc generator, that is, a generator where the polarity of the emf is always positive. The basic idea is to use a single split ring instead of two complete rings. The split ring is arranged so that, just as the emf is abo ...
... By a clever change to the rings and brushes of the ac generator, we can create a dc generator, that is, a generator where the polarity of the emf is always positive. The basic idea is to use a single split ring instead of two complete rings. The split ring is arranged so that, just as the emf is abo ...
07_Entanglement_in_nuclear_quadrupole_resonance_
... become the single most important theoretical paper in physics to appear since 1945; it was entitled On the Einstein Podolsky Rosen Paradox. • In 1964,John Bell showed that the predictions of quantum mechanics in the EPR thought experiment are significantly different from the predictions of a very br ...
... become the single most important theoretical paper in physics to appear since 1945; it was entitled On the Einstein Podolsky Rosen Paradox. • In 1964,John Bell showed that the predictions of quantum mechanics in the EPR thought experiment are significantly different from the predictions of a very br ...
Nature template - PC Word 97
... confinement respectively. The bias field is 0.75 G corresponding to a frequency of 2.1 MHz for a transition between the mJ = 1 and mJ = 0 states at the bottom of the trap. After evaporative cooling, we keep an RF knife on at constant frequency for 500 ms, then wait for 100 ms before switching off th ...
... confinement respectively. The bias field is 0.75 G corresponding to a frequency of 2.1 MHz for a transition between the mJ = 1 and mJ = 0 states at the bottom of the trap. After evaporative cooling, we keep an RF knife on at constant frequency for 500 ms, then wait for 100 ms before switching off th ...
Gauss` Law
... Define E2 to be equal to the magnitude of the electric field at r = 1.4 cm when the charge on the outer shell (q2) is equal to 1.1 μC. Define Eo to be equal to the magnitude of the electric field at r = 1.4 cm if the charge on the outer shell (q2) were changed to 0. Compare E2 and Eo. E2 = Eo Since ...
... Define E2 to be equal to the magnitude of the electric field at r = 1.4 cm when the charge on the outer shell (q2) is equal to 1.1 μC. Define Eo to be equal to the magnitude of the electric field at r = 1.4 cm if the charge on the outer shell (q2) were changed to 0. Compare E2 and Eo. E2 = Eo Since ...
