JEE Physics Magnetism and Matter Notes Set 03

Magnetisation, Magnetic Intensity and Susceptibility

Every atom has tiny current loops (orbiting and spinning electrons) and therefore magnetic moments. The net moment per unit volume of a sample is its magnetisation, \( M=\frac{m_{net}}{V} \), measured in A/m. Inside a long solenoid of n turns per metre carrying current i, the field without any core is \( B_0=\mu_0ni \). With a magnetised core the field becomes \( B=B_0+\mu_0M \).
The magnetic intensity \( H=\frac{B_0}{\mu_0}=ni \) depends only on the free current in the windings, whereas M describes how the material responds. For weak fields \( M=\chi H \), where the dimensionless susceptibility \( \chi \) is a property of the material. Then \( B=\mu_0(H+M)=\mu_0(1+\chi)H=\mu H \), with relative permeability \( \mu_r=1+\chi \) and permeability \( \mu=\mu_0\mu_r \). Only one of \( \chi \), \( \mu_r \) and \( \mu \) is independent.
A useful idea is the magnetising current \( i_m \): the extra current that, in the absence of the core, would produce the same B as the core does, so \( B=\mu_0n(i+i_m) \).

Diamagnetic, Paramagnetic and Ferromagnetic Substances

Diamagnetic: the atoms have zero net moment. An applied field changes the electron motion (by Lenz's law) so that a small moment opposite to the field appears; \( \chi \) is small and negative (of order \( 10^{-5} \)), \( \mu_r \) is slightly below 1, and the susceptibility does not depend on temperature. Such a body is weakly repelled, moves from strong to weak field regions, sets itself with its length perpendicular to the field, and its liquid level is depressed in the arm of a U-tube placed between the poles. Examples are bismuth, copper, gold, lead, water and nitrogen. Diamagnetism exists in all materials but is usually masked by other effects. A superconductor is a perfect diamagnet (\( \chi=-1 \), \( \mu_r=0 \)): it expels the field completely (Meissner effect).
Paramagnetic: atoms have permanent moments that thermal motion keeps randomly oriented; an applied field aligns them slightly. \( \chi \) is small and positive and obeys Curie's law \( \chi=\frac{C}{T} \) (equivalently \( M=C\frac{B_0}{T} \)) until saturation sets in at strong fields and low temperature. The body is weakly attracted, moves from weak to strong field regions, aligns parallel to the field, and its liquid rises in the U-tube arm. Examples: aluminium, sodium, calcium, platinum, oxygen at STP.
Ferromagnetic: neighbouring atomic moments lock together over regions called domains (about 1 mm across, with about \( 10^{11} \) atoms). In an applied field the domains aligned with the field grow and the others rotate towards it, so the substance is strongly magnetised: \( \chi \) is large and positive, \( \mu_r \) runs into hundreds or thousands, and the body is strongly attracted. Iron, cobalt, nickel and gadolinium are ferromagnetic. Above the Curie temperature the domain structure breaks up and the material becomes paramagnetic, with \( \chi=\frac{C}{T-T_c} \) for \( T>T_c \). The Curie temperature of iron is about 770 degrees C and that of nickel 358 degrees C.

Soft and Hard Magnetic Materials

In a hard ferromagnet (steel, alnico, lodestone) the magnetisation stays after the field is removed, so such materials make permanent magnets and compass needles. In a soft ferromagnet (soft iron) the magnetisation almost disappears when the field is removed, which suits electromagnets, transformer cores and other devices where the field is repeatedly reversed.
Hard materials have a high retentivity and a high coercivity; soft materials have a high permeability, low retentivity and low coercivity.

Hysteresis

When a ferromagnetic sample is taken through a full cycle of magnetising field, the magnetisation lags behind the field. Starting from an unmagnetised state, M rises with H to a saturation value. When H is reduced to zero, M does not return to zero; the value left is the retentivity (remanence). A reverse field must then be applied to bring M back to zero: its magnitude is the coercivity. Continuing the cycle traces a closed curve, the hysteresis loop.
The area enclosed by the loop (B plotted against H) equals the energy dissipated as heat per unit volume per cycle. A soft iron loop is narrow and tall, so the loss is small, which is why transformer and dynamo cores are made of it. A steel loop is wide, with large retentivity and coercivity, so it is used for permanent magnets. The heat generated per second in a core is (loop area per unit volume) times (volume) times (frequency).

Worked Examples

Worked Example 1. A solenoid with 1500 turns per metre carries 2.5 A and has a core of relative permeability 300. Find H, M and B.
Answer:
\( H=ni=1500\times2.5=3750 \) A/m, independent of the core.
\( M=(\mu_r-1)H=299\times3750=1.12\times10^6 \) A/m.
\( B=\mu_r\mu_0H=300\times4\pi\times10^{-7}\times3750=1.41 \) T.

Worked Example 2. A cubic domain of iron has an edge of 2 micrometre. Density of iron is 7.9 g/cm^3, molar mass 55 g/mol, and each atom has a dipole moment of \( 9.27\times10^{-24} \) A m^2. Find the number of atoms, the maximum dipole moment and the maximum magnetisation.
Answer:
Volume \( =8\times10^{-18} \) m^3 \( =8\times10^{-12} \) cm^3; mass \( =7.9\times8\times10^{-12}=6.32\times10^{-11} \) g.
Atoms: \( N=\frac{6.32\times10^{-11}}{55}\times6.022\times10^{23}=6.9\times10^{11} \).
\( m_{max}=6.92\times10^{11}\times9.27\times10^{-24}=6.4\times10^{-12} \) A m^2, and \( M_{max}=\frac{m_{max}}{V}=8.0\times10^5 \) A/m (the same as for any volume of iron, since all atoms are aligned).

Worked Example 3. The susceptibility of aluminium at 300 K is \( 2.3\times10^{-5} \). Estimate it at 150 K and at 600 K.
Answer:
Curie's law: \( \chi\propto\frac1T \).
At 150 K: \( \chi=2.3\times10^{-5}\times\frac{300}{150}=4.6\times10^{-5} \).
At 600 K: \( \chi=2.3\times10^{-5}\times\frac{300}{600}=1.15\times10^{-5} \).

Worked Example 4. Nickel (Curie temperature 631 K) is in its paramagnetic phase. Find the ratio of its susceptibility at 700 K to that at 800 K.
Answer:
For \( T>T_c \): \( \chi=\frac{C}{T-T_c} \).
\( \frac{\chi_{700}}{\chi_{800}}=\frac{800-631}{700-631}=\frac{169}{69}=2.45 \).

Worked Example 5. A soft iron core of volume \( 2\times10^{-3} \) m^3 is taken through 50 magnetisation cycles per second. The hysteresis loop (B against H) has an area of 400 J per m^3 per cycle. Find the power lost as heat.
Answer:
Energy per cycle \( =400\times2\times10^{-3}=0.8 \) J.
Power \( =0.8\times50=40 \) W.