Faraday's Law problem?
917169
L = μ(0) μ(r) N^2 A/l\n\nwhere\n \nμ(0) is the permeability of free space (4π × 10-7 henrys per metre) \nμ(r) is the relative permeability of the core (dimensionless) \nN is the number of turns. \nA is the cross sectional area of the coil in square metres. \nl is the length of the coil in metres\n\nHere we assume that μ(r) = 1, since the solenoid doesn't enclose anything like a metal core. Then\n\nL = [μ(0) N^2 (πR^2)] / l, where R is the radius of the solenoid\n\n= [(4π × 10-^7)(2920)^2(π(0.005)^2)] / (0.1)\n\n= 8.44 × 10-^3 henries
78681
If you could pulse the electricity in the wire, it may change the velocity of the wire.  Look up something called Magneto-hydrodynamics
233968
Biot-Savart law for a generalized current j can be written as:\n\n1) B(r) = μ0/4π rot ∫d³r' j(r')/|r - r'|\n\n\nthe scaled integral is a vector field, called the magnetic potential A:\n\nA(r) := μ0/4π ∫d³r' j(r')/|r - r'|\n\n\ntherewith you can write (1) as:\n\n2) B = rot A\n\n\nA is not unique. you can choose any scalar χ:\n\nA := A' + grad χ\n\nto satisfy equation (2), because for any χ\n\nrot grad χ == 0\n\n\nCoulomb Gauge: choose χ so that div A == 0\n\n\n\nMaxwell equations of Magnetostatics:\n\nM1) rot B == μ0 j\nM2) div B == 0\n\n\nequation (M1) with (2):\n\n<=> μ0 j == rot rot A\n\n<=> μ0 j == grad div A - ∆A \n\nwith Coulomb Gauge div A == 0\n\n<=> ∆A = - μ0 j\n\n\nThis equation is the fundamental problem of Magnetostatics:\n\n\ni ) given the current j in the volume V with the boundary condition ∂(V),\n\nii) solve ∆A == - μ0 j\n\n\nattend to the analogy to Electrostatics.\n\nPoisson Equation: ∆φ = - 1/ε0 ρ(r)\n\n\nThe problem is that the magnetic vector potential A is more abstract than its electrostatic scalar counterpart φ.
680309
staying inside a 'Faraday Cage' is a good first step.  Look up Faraday Cage in Wikipedia
696976
Let v1 = speed of the plug relative to a dot\nLet v2 = speed of the infinite pipe relative to a dot\nLet phi = angle of the pipe relative to some plane\nspeed of the pipe along the plane is v2 * Cos(phi)\nphi = 0 if we select a plane along the axis of the pipe\nTherefore the vector describing the speed of the pipe along the plane is v2*Cos(0) = v2\nThe plug speed relative to the plane is also found to be v1\nAn infinite pipe will act like a line that is very far away\nThe plug will act like a dot moving along that line if it is very far away too\nAssume that our origin is very far away from the pipe - if the pipe is infinite then space has to be very big and the assumption is correct for all 3-dimensional space.\nThe speed of the plug relative to the pipe is v1-v2\nv2 = 0 since if it was greater than 0 then it would have taken an infinite amount of energy to move it since it would have an infinite mass and no energy would have been left over to make the plug.\nSince the ends of the pipe are infinitely far away then the pipe must have come into existence after the plug\nThe guy who got the pipe wrapped around the plug would never have been able to do it if the plug was moving\nTherefore v1 = 0\nIf v1=0 and v2=0 then v1=v2\nSo in fact the plug is stopped relative to the pipe.
935341
1 sum the forces and set equal to zero. 2 forces, gravity and electric.\n\n2. conservation of energy. initially all kinetic, finally all potential. KE1=U2. U2 is electric potential. solve.\n\ngood luck
100023
The  Faraday constant divided by the Electron charge  =\n\n Avogadro's number
803176
That's very interesting.  I haven't heard of anyone with this problem, but it is interesting nonetheless.  I have some background in physics, and I'm not aware of any known scientific reason why that could be even remotely possible.  In my field, we often look for other more ordinary reasons why some phenomenon is occuring.  For example, it's possible that you've simply had bad luck with CD players.  Were all your CD players made in Bangladesh by any chance?  I've had some made there, and they all broke too, lol.  Car windows and radios malfunctioning could also likely be attributed to GM.  Trust me, I've been there too, lol.  Also, in the case of the car window, it could be your technique in operating the power windows.  Was there a child safety lock on the window that was locked at the time your mom was trying to roll them up?  I use the child safety lock in my car so my kids can't play with the power windows, and sometimes I forget it's on and I can't roll down the windows when I'm at the BK drive-thru, and I get so angry, I try screaming through the closed window, lol.  But seriously, when I get hungry I get a temper, and if I take it out on my kids one more time, I could loose custody, and I can't just keep blaming GM. \n\nAfter we explore the most common, earthly explanations for a particular phenomenon, we often find a perfectly acceptable answer, but sometimes we don't.  Science doesn't have all the answers yet.  The human body does have an electric charge, complete with an electromagnetic field.  I could see a slight possibility that if a person's electromagnetic field is exceptionally strong, it might interfere with electronics, although I think that's a long shot.  I've heard that sleeping on a bed of magnets, or wearing magnets in your clothing can correct electromagnetic imbalances, but there's no scientific proof to back that up.  I would say that there's nothing wrong with you that's causing electronics to malfunction, and it's more likely that you've just had bad luck with electronics.  But anything's possible.
193357
The generator consists mainly of a coil of wire wound over an iron core.This wire is connected to an external circuit where the current produced is either stored in capacitors or used to run a particular device. When the axle of the wire is rotated, magnetic flux linked with the wire changes and hence, according to Faraday's law, an electomotive force(emf) is induced in the wire as it is placed in between magnets having different polarities. Slip rings are present to allow the flow of the alternating current produced into an external circuit.
470161
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.
211344
Its in Wikipedia, the free encyclopedia, just type the "Faraday cage" in yahoo search you will find the 75% for your answer and rest you do your home work.
460932
Assume all the charges are concentrated at the center of the sphere.\n\nNow the problem is reduced to find the potential at a point which is at a distance equal to the radius of the sphere from the center of the sphere where all the charges are accumulated.
343981
In two charged plates, other than between the plates (and the edge effects), are there any field line? \nYES! other field lines are between the outer sides of the plates(depending on the charges of the plates and the convention used for the direction of the flow). To prove this, just connect the outer sides of the plates by a wire through an amper meter and you will see the meter moving.\n\nWhy is the electric field constant everywhere in the space between parallel plates? \nThey are not unless the area of both plates is inifinte and that the two plates are parallel.  In this case, the 1) direction of the field is the same every where since by definition the plates are parallel and 2) the absolute value is constant.  If not, there will be a surface current and the current would make the entire electric field (its magnitude) to be constant over the entire surface.
793702
Firstly,\nNaCl dissociates into Na+(aq) + Cl-(aq)\n1faraday of charge is required to produce 23 g Na from molten NaCl .\ntherefore production of 100000 g of Na will require charge = \n(1/23)*1000000 Faradays = 43478.26 Faradays = 43478.26 * 96500 C      [ as 1F = 96500 c]\n=4.1956 * 10^9 C\nBut, Q=IT \nso, T=Q/I = (4.1956 * 10^9)/300000 = 13.9855 * 10^3 secs = 3.88486 hrs.
737153
Starting from rest, the electron will have a kinetic energy of 25 electron volts (eV) after passing through the electric field.  One eV = 1.602*10^−19 J, and the kinetic energy (neglecting relativistic effects, which are not important here) is 0.5*m*s^2, where m is the rest mass of the electron, and v is its speed.\n\nWe have, therefore:\n\n25 eV * 1.602*10^−19 J/eV = 0.5*9.11*10^-31 kg * s^2\n\ns^2 = 8.79*10^12 (m/s)^2\n\ns = 2.96*10^6 m/s\n\nThe force on a particle with charge q, moving with vector velocity V in a magnetic field B is given by:\n\nF = qV x B\n\nwhere "x" means the vector cross product.\n\nIf, as it seems in this case, the velocity vector and magnetic field vector are perpendicular to one another, then  the magnitude of the force is simply given by the ordinary algebraic product of the charge, the magnitude of the velocity (i.e., the speed), and the magnitude of the magnetic field, so:\n\n|F| = q*|V|*|B|\n\n|F| = 1.60*10^-19 C * 2.96*10^6 m/s * 0.5 T \n\n|F| = 2.37*10^-13 N\n\n(It helps to know that 1 tesla = 1 kg/(sec*coulomb) here).\n\n \nThe force vector is oriented perpendicular to both the velocity and the magnetic field.  However, there are still two possible directions that satisfy this condition.  The ambiguity can be resolved by using the "right hand rule":  point your index finger along the direction of the velocity vector, and your middle finger along the direction of the magnetic field.  Your thumb will then point in the direction of the force vector.  You'll have to figure out how this applies to what I assume was the figure provided with your problem.\n\n\n\nTo determine the radius of the electron's orbit, we equate the magnetic force with the centripital force:\n\nq*|V|*|B| = (m*|v|^2)/r\n\nr = (m*|v|^2)/(q*|V|*|B|)\n\nr = (m*|v|)/(q*|B|)\n\nr = (9.11*10^-31 kg * 2.96*10^6 m/s)/(1.60*10^-19 C * 0.5 T)\n\nr = 3.38*10^-5 m
288780
From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
942587
poorcocoboiboi is correct, however the electric field is still there even after the charge is moving. In fact it is the moving electric field that produces the magnetic field.