x x| ------->\nx x| x x x x\nx x| x x x x \nx x| x x x x\nx x| x x x x\nconsider a wire of length L moving with a constant velocity V\ntowards the right\nIt cuts a uniform magnetic field of induction B pointing into the monitor screen \nResistance of wire is R\nfind the magnitude and direction of both the EMF and current in the wire
213065
The above answer is wrong, that equation is for the force on the wire!\n\nFor the magnitude of the induced EMF, you need to use BLV.\nThen Use V=IR to get the current.\n\nThe directions are a bit more tricky - you have to use the Left hand rule or the corkscrew rule to do this bit.\nJust remember than to obey Lenz's Law, the current must flow in such a way that the force on the wire given by BIL must oppose the direction of motion of the wire.\n\nSo, to put it simply, the curent has to be up in the diagram you drew above.\n\nBy the way, in order for current to flow, you must consider that this wire is part of a complete circuit.
214633
You've actually got several concepts and questions merged together.\n\nWhen you ask "how does the aerial setup the waves from only the ac signal?"  I assume you're asking how any DC components of the signal get filtered out?\n\nA transverse electromagnetic wave involves both a VARYING electric field and a VARYING magnetic field, appearing at right angles to each other and to the direction of travel of the wave.\n\nAny DC component has no varying component, therefore, no wave is generated.  Although DC flowing through a wire creates electric and magnetic fields, no wave is generated because there is no variance in the fields.  \n\nThere are other electronics concepts that require the changing electromagnetic fields of AC to work.  A transformer uses AC, not DC.  A capacitor will block DC signals, but allow AC signals to pass.\n\nAlso, as you indicated, an efficient antenna has a matched impedance to the signal source in order to maximize the amount of engergy propagated by the antenna in the desired direction(s).
202011
If a gets one coil, there are four ways.\n\na(1), b(3), c(0)\na(1), b(2), c(1)\na(1), b(1), c(2)\na(1), b(0), c(3)
491970
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
431386
T1=m1*a, no matter are there m2 & m3 or not.\nT2=(m1+m2)*a, no matter whether m1 & m2 are connected with a string or are in one lump, the presence of m3 being unimportant too.
140986
See the other postings of this question.  The equation is\n\nhttp://members.aol.com/logan20/elecspon.html
393809
this question was on this yr's midyear exam, ill give u a clue.\nyou need to seperate the forces acting on the wire into two types \n1) acting horizontally (the force to the left equals the force to the right)\n2)acting vertically (the force to the bottom {or the normal force = 20 *9.8} equals the force to the top)\n\nequate them both and you should get the answer
297276
(m)\n\ngoto the following website for your question with neat diagrams\n\nhttp://hyperphysics.phy-astr.gsu.edu/HBASE/electric/gaulaw.html
87683
Electromagnetic induction is basically the generation of an electric current in a wire due to passing through lines of magnetic flux, or changing flux.  This is the underlying principle in virtually all electromechanical generators in use today.\n\nProbably the easiest way to store the electrical energy would be in chemical form, such as a battery.  You could also store smaller amounts of energy in capacitors as well.\n\nWiring is simple, take a coil of wire and move it through a magnetic field.  The field can be from a permanent magnet, or from another coil of wire that has an electric current passing through it, creating a magnetic field.
149783
1.  not enough info\n2.  conduction?\n3.  conductor\n4.  resistance
420968
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
153311
Gee .. been a while.  If I remember correctly the magnetic field is perpendicular to the electrical field.  Thus around the two wires the magnetic fields move in the same direction and combine, producing a net attractive force.\n\nIf the currents go the opposite direction, the forces repel.\n\nSee the link below for a more detailed explanation and pictures:
284601
Since Magnetic field (B) = [K I(dl x r)]/r^3\n\nwhere I = current in the conductor\n      r = distance of the pt. from the conductor where u  want to find the magnetic field\n\nHere dl x r = vector product of small part of L i.e dl and r\n\nSee here that Magnetic field is directly proportional to Current....therefore when we increase the current by a factor of 4, magnetic field will also increase by a factor of 4.
121182
Use the node voltage method. Label each node with a voltage, then apply Kirchoff's current law at as many nodes as you need to generate enough equations to solve for any unknowns you have introduced.\n\nLet the node at the bottom of the circuit be ground, and then try to figure out the voltages at the other nodes relative to ground. Clearly the voltage at the node on the top will be 14V (I'll just call this V1) but no other voltages are apparent. So call the voltage at the node between R1, R3, and R4 Vx and the voltage at the node between R2, R3, and R5 Vy.\n\nNow that we have labeled all the node voltages, you can figure out every single current in the circuit in terms of V1, Vx, and Vy (using Ohms Law, the voltage drop across a resistor is equal to the resistance times the current in the direction of the voltage drop). For example, we know the current going down through R1 is (V1-Vx)/R1, the current going right through R3 is (Vx-Vy)/R3, etc. Now we just have to solve for the unknowns Vx and Vy.\n\nKirchoff's current law (KCL) says that the sum of the currents leaving a node is equal to zero. If we apply this law to the node with voltage Vx and the node with voltage Vy, we'll get a pair of equations that we can use to solve for Vx and Vy.\n\nApplying KCL at the node with voltage Vx should give you the following equation:\n\n(Vx-V1)/R1 + (Vx-Vy)/R3 + Vx/R4 = 0\n\nThe first term is the current going up through R1 (we already figured out the current going down through R1, but we want currents that are leaving node Vx so we had to flip the sign), the second term is the current going right through R3, and the final term is the current going down through R4.\n\nI'll leave it to you to figure out the equation for node Vy. Once you have that you'll have two equations and two unknowns and you can solve for everything. Once you have Vx and Vy, you can easily figure out a numerical value for any current in the circuit.\n\nBTW, the node-voltage method is pretty powerful and can solve most linear circuits. Check out the link below, it was written by an old EE prof of mine who is very thorough.
31044
If you could pulse the electricity in the wire, it may change the velocity of the wire.  Look up something called Magneto-hydrodynamics
198294
If you get an A with this work please share it with me. If you get an F , forget me.\n\n  \n  Magnetism are 2 sides (+ and -) (like in some relations). Every compass will go to the North because of magentic fields, those fileds are kind of circles like the air waves of radio or Tv.\n\n  2. Electromagentic is a toold like iron that needs electricity to work and work as a magnet, it is important a battery or other source of low voltage. Electromagentics are used to attract old cars, electromagentics are used for people who lost some things on the sand (those kind of electromagnetics are the ones with one circle in the end where the person is attracting something).\n\n 3: Electromagentic needs a switch like light, turn on and off, you let the energy flow and you get the magnetic field or you deny that flow, on the other hand the magnet is always "on", there are magnets where both sides are positive or negative, most of them are positive and negative like in electricity you need both to have electricity.\n\n 4:For instance the motor of one car, inside the motor you will see a lot of pieces and is kind of interesting because thin cables are allowing the flow of electricity, so with that in mind the motor is moving, produces energy then
484084
1. Xc = 1/2πfC\nSo 2πfC = 1/Xc\nSo C = 1/2πfXc\n\n2. ib = ibq + √2is\nSo √ = ib - ibq\nSo 2is = (ib - ibq)²\nie is = ½(ib - ibq)²\n\n3. Q = XL/R\nSo XL = QR\n\n4. Lt = L1 + L2 - 2Lm\nSo 2Lm = L1 + L2 - Lt\nie Lm = ½(L1 + L2 - Lt)\n\n5. R1=i2 R2 / i2\n??????? So R1 = R2\n\n6. Es = E1 + E2 + E3\nSo E2 = Es - E1 - E3\n\n7. R1 / Rx = R2 / R3\nSo Rx / R1 = R3 / R2\nSo Rx = R1 * R3/R2\n\n8. y = Mx + b\nSo Mx = y - b\nSo M = (y - b)/x\n\n9. i2=i1 R1 / R2\nSo i1 R1 = i2R2\nie i1 = i2 R2 / R1\n\nI would be extremely concerned about your progress if you find this simple algebra a serious chore!!!!!! (Especially if you are learning the electrical trade)