Solutions to Some Problems in Magnetism

  1. A long bar magnet has pole area A. An identical magnet is brought up close to the first so that a pair of opposite poles are parallel and almost touching. The magnetic field between them is B. Find the force of attraction.
  2. A long thin cylindrical bar magnet, length L and radius r, has magnetic field Bo at its poles. Find its dipole moment. (Note: if this were an air-core coil, the dipole moment would be NIp r2.)
  3. A cylindrical magnet has radius r, length 4r, and magnetic field Bo at the poles. Find the magnetic field a distance 2r from the end on the axis of the cylinder.
  4. With the magnet in problem 2, find the field at a point that forms the right angle corner of a 3-4-5 triangle with the magnet as the hypotenuse.

Solutions

    1. Assume constant B over area A; B small elsewhere (like the parallel-plate capacitor assumption). Energy/volume in B field is B2/2m o. Pull them apart a tiny distance x, doing work Fx, and increasing the energy by (B2/2m o)Ax. Equate these, cancel x and get F=AB2/2m o. Another way is to figure the force between capacitor plates in terms of E, then replace E with B and e o with 1/m o.
    2. Assume that it has the same magnetic moment as a long coil that is producing the same Bo at its end. For a coil, Bo = (m oNI/2L)(cosa +cosb ) where a and b are the angles between the axis and the corners of the coil. In this case approximately zero and 90o, so Bo = m oNI/2L. Hence NI = 2BoL/m o. The magnetic moment m B is NI times area, so m B = 2LBo pr2/m o.
    3. Assume that the field is the same as the field of coil that has field Bo at its end. Use the formula above, B = (m oNI/2L)(cosa +cosb ). In this case at the point 2r from the end the cosines are 6r/(36r2 + r2)1/2 and -2r/(4r2 + r2)1/2. Do the math and show (cosa +cosb ) = 0.092. For Bo, the cosines are 4r/(16r2 + r2)1/2 and 0. Show Bo = 0.97m oNI/2L, so replace m oNI/2L with Bo/0.97 in the equation for B, and we have B=Bo0.092/0.97 = 0.095Bo
    4. Assume that it acts like the electric field due to oppositely charged spheres of radius r, long distance L apart, so Eo = kq/r2 at a sphere surface. Draw the picture and you will see that there are two fields at right angles, toward the -q and away from the +q. One is kq/(0.6L)2 and the other is kq/(0.8L)2 . Take the square root of the sum of the squares, and replace kq with Eor2, and get E = 3.2Eor2/L2 . The tangent of the angle is 0.36/0.64 or the inverse of this, and we get 32o or 58o from the line to the negative side, depending on whether that is the closer or farther one. Now replace Eo with Bo and E with B.

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