Data Table One Angle of Inclination ( ф ) Deflection current (l45) - L-60- 201.5mA Turn on the ...
nt FULL SCREEN PRINTER VERSION NEXT Cnapter 41, ProDiem b2 A smail compass is held horizontally, the center of its needle a distance of 0.240 m directly north of a long wire that is perpendicular to the earth's surface. When there is no current in the wire, the compass needle points due north, which is the direction of the horizontal component of the earth's magnetic field at that location. This component is paraliel to the earth's surface. When the current...
A long current-carrying wire, oriented North-South, lies on a table (it is connected to batteries which are not shown). A compass lies on top of the wire, with the compass needle about 3 mm above the wire. With the current running, the compass deflects 10 degrees to the West. At this location, the horizontal component of the Earth's magnetic field is about 2e-5 tesla. What is the magnitude of the magnetic field at location A, on the table top, a...
This is my experiment about Earth magnetic field induction by tangensgalvonometer. I must additional conclusion for my experimental values and Bo=1.68x10^-5 T. Could you compare Bo=1.68x10^-5 and my experimental values. experiment did same city which is Bo=1.68x10^-5 I need conclusion about this case. thank you. Results: Bo (1-1 )-7.66x10 Be(l 2-0107 Be (1-3 )-1.72x10 · Be (1-3, 2 )" f,79x16 Bo (Rī9小1.GC-er「 Measurement of the horizontal component of Earth's magnetic field induction by tangensgalvanometer In each point the Earth's magnetic...
3 of Review Cone Part B What is the necessary current? Express your answer to two significant figures and include the appropriate units. Bats are capable of navigating using the earth's foldha plus for an animal that may ily great distances from its roost at night. It, while sleeping during the day, bats are exposed to a field of a similar magnitude but different direction than the earth's field, they are more likely to lose their way during their next...
The size of the magnetic force on a straight wire of length L carrying current I in a uniform magnetic field with strength B is F=ILBsin(?). Here ? is the angle between the direction of the current (along the wire) and the direction of the magnetic field. Hence Bsin(?) refers to the component of the magnetic field that is perpendicular to the wire, B?. Thus this equation can also be written as F=ILB?. The direction of the magnetic force on...
Clearly state your solution path. Partial credit will only be given for clear steps. Units must be provided or points will be subtracted. Clearly label the end result, use double lines to underscore or box the final result. Note the equation sheet is on the back. (10 pt) An electron enters a region between two capacitor plates with equal and opposite charges. The plates are L=0.1 m by 0.05 m and the gap between the plates is h=0.001 m. During...
Table 1: Data for Solenoid 1. Voltage IV) Current A o 0 06 10 031 0 011 020 099 07 2.71 35 0.96 1.25 19 25 5.7 Taille 2. Data for Solenoid 2. Current(A) 2005 003 0.075 0.1 125 9.15 0.2 0225 .244 1.146 289 382 47 5.78 7.68 821 Table 3. Data for field as a function of distance from center of Solenoid (Current A Distance center of solenoid 55 ledge of 812.5 cm from deel 30.505 cm from...