Re: Is there an electromagnet that produces a very narrow beam (1/2in.) and

A current loop in a uniform static magnetic field behaves like a magnetic dipole (a small magnet). The torque exerted on it by uniform static magnetic field is
T = m x B
Here "x" denotes vector product. T, m and B are all vectors. m is the magnetic moment of the current loop. Its value is m=IA, where I is the current, A is the area of the loop. Its direction is normal to the loop plane and obeys the right-hand rule in the relation to the direction of the current. B is the magnetic flux density.

From this equation we can see the torque produced by a magnet is related to the area and current of the loop it is supposed to rotate. For a 1,000 turns 0.5 inch x 0.5 inch square loop coil with a 10A current, its magnetic moment is 1,000* 1A*(0.0127m)²=0.16 A*m². To exert a 100 ft-lbs (=100*0.3048m*4.448N=136 N*m) torque on such a coil, a magnetic flux density of 136 N*m/(0.16A*m²)=850 T is required.

Generally a magnetic field above 20 Tesla is regarded as high field,  power supply of several megawatts and water-cooling system are needed to maintain such an electromagnet.

Here is an example of electromagnet with design procedure. It weighs 15kg and generates a magnetic flux density of  1174 gauss (0.1174 Tesla).

To decrease the weight of electromagnet, thinner wires or fewer turns are required, however, that means larger current is needed. The heat generated by the magnet will become the biggest problem.

In a summary, it seems impossible to design a electromagnet which can produce a 100 ft-lbs of torque on a usual coil and weighs less than 2 lbs. However, by cooling the whole system to a very low temperature and make the coils superconductive, very large currents can be used. In this situation such a design may be possible.