# Electromagnetism- current carrying conductor and permanent magnet

Hi.
I'm doing an experiment and i'm kinda stuck.

I know that a current produces a magnetic field. In this experiment there is a current carrying wire (copper) and a permanent magnet suspended above it. Deflection is being measured as the current is varied.

I know that both magnetic fields will interact and that a force would be exerted on the permanent magnet (as it is suspended and the wire statioonary). The force would therefore be dependant on the magnitude of the current as well as the magnetic flux density.

So the eqn explaining this is F= BIl? I could be wrong. Does this then mean that the permanent magnet cuts the magnetic field of the wire? As the wire would be stationary, hence parallel to the magnetic field of the permanent magnet? If so, would the angle made by the magnet cause the Force to be determined by F= BIl (sin theta)?

jambaugh
Gold Member
Without specific analysis given the shape of the magnet and the coil of wire you will not get an exact formula for the force (I presume you actually mean torque?). It should however be proportional to the current and roughly to the sine angle.

There will be smaller higher order contributions to the angle dependence ( sin(2theta) and higher terms) since the magnet is not a perfect dipole source and the coil is not producing a uniform B field around it. But torque proportional to I*B*sin(theta) will be a quite close approximation.

jambaugh
Gold Member
Also as to understanding the nature of the force, there is an energy density proportional to the square of the total B field
$$\vec{B} = \vec{B}_{mag} + \vec{B}_{coil}$$

Total energy is an integral:
$$E = \int d^3 x B^2(x) = \int d^3 x [B_{mag}^2 + B_{coil}^2 + 2\vec{B}_{coil}\cdot \vec{B}_{mag}]$$
You get the energy of the coil alone plus the energy of the magnet alone plus the cross terms. The cross terms will add up to a minimum energy when the to fields cancel as much as possible so the magnet will turn to oppose the field generated by the coil and this is strongest where the change in energy has a highest rate. Typically when they are at right angles the torque is maximized.

the magnet is a bar magnet and the current carrying wire is actually a straight wire...would that make any difference?

jambaugh
Gold Member
the magnet is a bar magnet and the current carrying wire is actually a straight wire...would that make any difference?

Some. It depends on the distance from the wire relative to the length of the bar magnet. The farther away the closer to a pure sin(theta) form but of course also the weaker the force.

What's happening there is the magnitude force will be affected also to the distance of the ends of the bar magnet from the wire which also is a function of the angle. The farther the distance the more constant that function will be and the closer to pure sin(theta) dependence.

Also the angle we're speaking of is the angle of the bar magnet from general direction of the wire's B field not from the line of the wire which is 90deg off.

Also note you still have a loop of wire since you have a circuit.

Thank you very much. This was really helpful

I have dangled neodymium-iron-boron magnets on a thread in the Earth's magnetic field to determine their polarity. If there were a north-south wire with a current, the east-west magnetic field from the wire would put an additional torque on the magnet. So a 1 Gauss B field fom the wire would rotate the magnet about 45 degrees.
Bob S