A physics question on magnetism?
920208
If the current is flowing into the paper from up to down, then using the right-hand rule, we determine that the magnetic field lines are in concentric cirles rotating in a clockwise direction.
871516
mass of the object\nforce of the magnet in a magnetic field\nthe object's own interference with the magnet
674108
The poles of any magnetized item will be at the extreme opposite points...in other words the positive will be as far as it can get from the negative.  In the case of a sphere this would be "top" and "bottom" not core and perimeter.  (Think of the Earth's magnetic poles!).
920912
Magnetization of materials\n\nFerromagnetic materials can be magnetized in the following ways:\n\n* Placing the item in an external magnetic field will result in the item retaining some of the magnetism on removal. Vibration has been shown to increase the effect. Ferrous materials aligned with the earth's magnetic field and which are subject to vibration (e.g. frame of a conveyor) have been shown to acquire significant residual magnetism.\n    \n* Placing the item in a solenoid with a direct current passing through it.\n    \n* Stroking - An existing magnet is moved from one end of the item to the other repeatedly in the same direction.\n    \n* Placing a steel bar in a magnetic field, then heating it to a high temperature and then finally hammering it as it cools. This can be done by laying the magnet in a North-South direction in the Earth's magnetic field. In this case, the magnet is not very strong but the effect is permanent.\nMagnetization of materials\n\nFerromagnetic materials can be magnetized in the following ways:\n\n* Placing the item in an external magnetic field will result in the item retaining some of the magnetism on removal. Vibration has been shown to increase the effect. Ferrous materials aligned with the earth's magnetic field and which are subject to vibration (e.g. frame of a conveyor) have been shown to acquire significant residual magnetism.\n    \n* Placing the item in a solenoid with a direct current passing through it.\n    \n* Stroking - An existing magnet is moved from one end of the item to the other repeatedly in the same direction.\n    \n* Placing a steel bar in a magnetic field, then heating it to a high temperature and then finally hammering it as it cools. This can be done by laying the magnet in a North-South direction in the Earth's magnetic field. In this case, the magnet is not very strong but the effect is permanent.
128833
Three types of magnets:\n\n-Ferromagnets: Magnets which retain a magnetic field after having been magnetised in a magnetic field.\n\n-Paramagnets: Magnets that only exert a magnetic field when they're in another magnetic field.  They support the magnetic field they're in.  Usually their fields are quite weak compared to the external fields.\n\n-Dimagnets: Magnets that exert a magnetic field when they're in another magnetic field but opposing the field.  (Think: *DI*ametrically opposite).  Again, field intensity created by these magnets typically is very small compared to the external field.\n\nAdditionally, many dimagnetic and paramagnetic materials may become ferromagnetic if they are sufficiently cooled.\n\nPermanent magnets are most often ferromagnetic and will retain their magnetism for a long time even in the absense of a magnetic field.  Over time, however, thermal effects will slowly decrease field intensity, but that takes a long time for permanent magnets.
1074720
none of the answers is correct , actually magnetic lines of force will be acting in anticlockwise direction.
85373
Magnetic force........
130907
Most magnets are formed by moving electrons. Each moving electron creates a small magnetic field. If lots of electrons are moving along the same path, they create a bigger magnetic field. Because of this you can make a magnet by using electric current. If you wrap a coil of wire around a piece of iron (like a big nail) and then put electricity through it using a battery, you will create a magnetic field. This is what we call an "electromagnet." Remember, only use batteries if you try this, never use a wall outlet--this is much too dangerous!!
598602
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.
88063
The previous correspondent confirmed that only a moving charge is affected by a magnetic field, but didn't actually address the question of *why* that is the case.  I will attempt to do so.  First, realize that electric fields  and magnetic fields are, under the theory of special relativity, two interrelated aspects of the same thing. Like momentum and energy, or space and time, they vary, depending on the reference frame of the observer. The values for the electric and magnetic fields in one reference frame can be transformed to the values for another reference frame by using the Lorentz transformation.\n\nNow, imagine a charge at rest in a given reference frame.  It exerts a force on other charges given by Coulombs' law.  The amount of force it exerts on those other charges depends only on the amount of charge each of our charges has, and the distance between them.  Another way of looking at this is to say our charge sets up an electric field, and the force on the other charges depends only on the strength of the field and the amount of the other charges.\n\nNow imagine that we are looking at this charge from the point of view of a reference frame that is moving with respect to the charge, so that to us the charge appears to be moving.  We take the formula for the force exerted by the charge, apply the Lorentz transformations which we have to use to relate quantities in different reference frames, and lo and behold, we not only get a force that depends on the amount of the other charges, we get a term that depends on the relative velocity of the charges.  What's more, the force of this new term is at right angles to the relative motion.  We identify this new part of the force, the part that depends on relative velocity of the charges, as the magnetic force.  Hence, the magnetic force depends on the relative motion of the charges.
1045884
http://en.wikipedia.org/wiki/Magnets\n\ngo there if u wanna find out, i only know simple version...
1135256
Conduction?  I'm not aware of magnets having such a property.
512643
ferromagnet materials will lose their magnetism if heated above a point known as the Curie temperature. \nExposure to an opposing magnetic field will "knock it down", but not demagnetize it completely.
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.
539149
A bit rusty, but here it goes - I beleive they subtract fro m the net magnetization. If some of the ions are anti-aligned it is called ferrimagnetic. If the ions anti-align completely it is called antiferrmagnetic.
25598
The energy isn't destroyed, but it is reorganized into different forms.  Think of it this way: Imagine you have a bunch of fridge magnets with words on them, and you arrange them into a brilliant poem.  When you come back later, someone else has rearranged the magnets.  All of the magnets have been "conserved"-- they're still there, and they're just as magnetty as ever-- but the particular shape that they were in when they were your brilliant poem is unrecoverable.  Sorry. <3
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