Exercise practice: Magnetism and Matter

4.1 Exercise 5.1: torque on a bar magnet

A short bar magnet is placed with its axis at 30 degrees to a uniform field of 0.25 T and feels a torque of 4.5 x 10 to the power -2 J. Use tau = mB sin theta. So m = tau / (B sin theta) = 4.5 x 10^-2 / (0.25 x sin 30 degrees) = 4.5 x 10^-2 / (0.25 x 0.5) = 0.36 A m squared.

4.2 Exercise 5.2: stable and unstable positions

The magnet has m = 0.32 A m squared and the field is 0.15 T. The stable position has m parallel to B, so the potential energy is -mB = -0.048 J. The unstable position has m antiparallel to B, so the energy is +0.048 J. The difference is 0.096 J, which is the work needed to turn the magnet from one position to the other.

4.3 Exercise 5.3: a solenoid as a magnet

A closely wound solenoid has 800 turns, an area of 2.5 x 10 to the power -4 m squared, and carries 3.0 A. Its magnetic moment is m = NIA = 800 x 3.0 x 2.5 x 10^-4 = 0.60 A m squared. The sense is given by the right hand rule: looking at one end, if the current is clockwise that face is the south pole and the moment points into the solenoid away from the observer. The solenoid behaves like a bar magnet of the same moment, and its field at a distant axial point follows the short dipole formula.

4.4 Exercise 5.4: torque on the solenoid

If the solenoid of the previous exercise is free to turn about a vertical axis and a horizontal field of 0.25 T is applied, the torque when its axis makes 30 degrees with the field is tau = mB sin theta = 0.60 x 0.25 x 0.5 = 0.075 N m.

4.5 Earth's magnetic field questions

The horizontal component of the earth's field at a place is B_H and the angle of dip is delta. The vertical component is B_V = B_H tan delta, and the total field is B = B_H / cos delta. If a place has a dip of 30 degrees and a horizontal component of 0.22 gauss, then B_V = 0.22 x tan 30 degrees = 0.127 gauss and B = 0.22 / cos 30 degrees = 0.254 gauss. The same relations answer the reverse question, in which the total field and the dip are given and the components are required.

A dip circle at the magnetic equator has its needle horizontal, because the field there is entirely horizontal, so the dip is zero. At the magnetic poles the field is vertical, so the needle stands vertical and the dip is 90 degrees. Between the two the dip changes gradually, and the values are equal at places on the same magnetic latitude.

4.6 Material questions

A material has a susceptibility of -0.5 x 10 to the power -5. The negative sign makes it diamagnetic, and its relative permeability is 1 + chi = 0.999995, just below one. If a material has chi = 2 x 10 to the power -3, it is paramagnetic, with a relative permeability of 1.002. A ferromagnet would need a susceptibility of the order of 1000 or more.

The Curie law question is a ratio problem. If the susceptibility of a paramagnet at 300 K is chi, then at 600 K it becomes chi/2, because chi is proportional to 1/T. If the material is ferromagnetic at room temperature and heated above its Curie temperature, its domains break up and it becomes paramagnetic, so its susceptibility falls sharply and it is no longer strongly attracted.

4.7 Permanent magnet against electromagnet

An electromagnet is made from soft iron because its narrow hysteresis loop means little energy is lost in each magnetisation cycle and it loses its magnetism as soon as the current stops. A permanent magnet is made from steel or alnico because its wide loop gives high retentivity and high coercivity, so it keeps a strong field after being magnetised and resists demagnetisation by stray fields or by a modest reverse field.