Magnetic Effects of Current and Magnetism • Topic 3 of 3

Magnetism & Matter

The bar magnet as a dipole. A bar magnet behaves like a magnetic dipole with two equal and opposite poles. Its magnetic dipole moment is $\vec{m}=m_p\,(2\vec{l})$, a vector pointing from the south to the north pole; its SI unit is $\text{A m}^2$. As with electric dipoles, isolated magnetic poles (monopoles) have never been found — cutting a magnet always yields smaller complete dipoles.

Field of a magnetic dipole. Treating the bar magnet like an electric dipole (with $\frac{\mu_0}{4\pi}$ replacing $\frac{1}{4\pi\varepsilon_0}$), the field at a point on the axial line at distance $r$ is $B_{axial}=\frac{\mu_0}{4\pi}\frac{2m}{r^3}$, and on the equatorial line it is $B_{eq}=\frac{\mu_0}{4\pi}\frac{m}{r^3}$. The axial field is therefore twice the equatorial field and points along $\vec{m}$, while the equatorial field opposes $\vec{m}$.

Torque and potential energy. In a uniform field a dipole feels a torque $\vec{\tau}=\vec{m}\times\vec{B}$ (magnitude $mB\sin\theta$) and has potential energy $U=-\vec{m}\cdot\vec{B}=-mB\cos\theta$. It is in stable equilibrium when aligned with the field ($\theta=0$) and unstable when anti-aligned ($\theta=180^\circ$).

The Earth's magnetism. The Earth behaves like a giant bar magnet tilted from its rotation axis, with its magnetic south near the geographic north. Three elements of terrestrial magnetism describe the field at a place:

  • Declination ($D$): the angle between geographic north and magnetic north (the horizontal direction the compass points).
  • Dip or inclination ($I$): the angle the total field makes with the horizontal. At the magnetic equator $I=0^\circ$; at the magnetic poles $I=90^\circ$.
  • Horizontal component ($B_H$): $B_H=B\cos I$ and the vertical component $B_V=B\sin I$, so $\tan I=\frac{B_V}{B_H}$.

Magnetic intensity and material quantities. Inside a material the field is $\vec{B}=\mu_0(\vec{H}+\vec{M})$, where $\vec{H}$ is the magnetising field and $\vec{M}$ the magnetisation (moment per unit volume). The susceptibility $\chi=\frac{M}{H}$ and the relative permeability $\mu_r=1+\chi$ classify materials.

Three classes of magnetic materials.

  • Diamagnetic (e.g. bismuth, copper, water): weakly repelled by a magnet; $\chi$ is small and negative; $\mu_r$ slightly less than 1. They move from strong to weak field regions.
  • Paramagnetic (e.g. aluminium, sodium, oxygen): weakly attracted; $\chi$ is small and positive; $\mu_r$ slightly more than 1. Magnetisation follows Curie's law $\chi\propto\frac{1}{T}$.
  • Ferromagnetic (e.g. iron, cobalt, nickel): strongly attracted; $\chi$ is large and positive; $\mu_r$ very large. Above the Curie temperature they become paramagnetic.

Hysteresis. When a ferromagnetic sample is taken through a full cycle of magnetising field, the $B$-$H$ graph forms a closed loop called the hysteresis loop. The lag of $B$ behind $H$ is hysteresis. The field $B$ left when $H=0$ is the retentivity, and the reverse $H$ needed to make $B=0$ is the coercivity. The loop area equals the energy lost per cycle. Soft ferromagnets (soft iron) have a thin loop, low coercivity and small loss — ideal for electromagnets and transformer cores; hard ferromagnets (steel, alnico) have a wide loop and high retentivity — ideal for permanent magnets.

Torque on a current-carrying rectangular loop in a magnetic field: tau equals N I A B sin thetaBBADBCForce outForce innormal ntau = N I A B sin(theta)
1
Worked Example
A bar magnet of pole strength 20 A m has its poles 0.10 m apart. Find its magnetic dipole moment.
Solution
  1. Step 1: Magnetic moment $m=m_p\times(2l)$, where $2l$ is the separation of the poles.
  2. Step 2: Substitute $m_p=20$ A m, $2l=0.10$ m.
  3. Step 3: $m=20\times0.10=2.0$ A m$^2$.

Answer: $m=2.0\,\text{A m}^2$, directed from S to N.

2
Worked Example
Find the magnetic field on the axial line at 0.20 m from a short bar magnet of moment 0.5 A m$^2$.
Solution
  1. Step 1: Axial field $B_{axial}=\frac{\mu_0}{4\pi}\frac{2m}{r^3}$.
  2. Step 2: $\frac{\mu_0}{4\pi}=10^{-7}$, $m=0.5$, $r^3=(0.20)^3=8\times10^{-3}$.
  3. Step 3: $B=10^{-7}\times\frac{2\times0.5}{8\times10^{-3}}=10^{-7}\times125=1.25\times10^{-5}$ T.

Answer: $B_{axial}=1.25\times10^{-5}\,\text{T}$, directed along $\vec{m}$.

3
Worked Example
A magnetic needle of moment 0.8 A m$^2$ is held at $30^\circ$ to a uniform field of 0.4 T. Find the torque on it.
Solution
  1. Step 1: Torque $\tau=mB\sin\theta$.
  2. Step 2: Substitute $m=0.8$, $B=0.4$, $\theta=30^\circ$ ($\sin30^\circ=0.5$).
  3. Step 3: $\tau=0.8\times0.4\times0.5=0.16$ N m.

Answer: $\tau=0.16\,\text{N m}$, tending to align the needle with the field.

4
Worked Example
At a place the horizontal component of the Earth's field is $0.34\,\text{G}$ and the angle of dip is $60^\circ$. Find the total field.
Solution
  1. Step 1: The horizontal component is $B_H=B\cos I$, so $B=\frac{B_H}{\cos I}$.
  2. Step 2: $\cos60^\circ=0.5$.
  3. Step 3: $B=\frac{0.34}{0.5}=0.68$ G.

Answer: $B=0.68\,\text{G}=6.8\times10^{-5}\,\text{T}$.

5
Worked Example
Distinguish between diamagnetic, paramagnetic and ferromagnetic materials on the basis of susceptibility.
Solution
  1. Step 1: Diamagnetic: $\chi$ small and negative ($\mu_r<1$); weakly repelled (e.g. bismuth, copper).
  2. Step 2: Paramagnetic: $\chi$ small and positive ($\mu_r>1$); weakly attracted; $\chi\propto1/T$ (e.g. aluminium).
  3. Step 3: Ferromagnetic: $\chi$ large and positive ($\mu_r\gg1$); strongly attracted; becomes paramagnetic above the Curie temperature (e.g. iron).

Answer: Diamagnetic $\chi<0$, paramagnetic small $\chi>0$, ferromagnetic large $\chi>0$.

6
Worked Example
Why is soft iron used for electromagnets and transformer cores while steel is used for permanent magnets? Explain using hysteresis.
Solution
  1. Step 1: The energy lost per magnetisation cycle equals the area of the hysteresis loop.
  2. Step 2: Soft iron has a thin loop with low coercivity and low retentivity, so it magnetises and demagnetises easily with little energy loss.
  3. Step 3: This makes soft iron ideal for electromagnets and transformer cores (low loss, switchable).
  4. Step 4: Steel has a wide loop with high retentivity and coercivity, so it retains magnetism strongly — ideal for permanent magnets.

Answer: Soft iron (thin loop, low loss) suits electromagnets and cores; steel (wide loop, high retentivity) suits permanent magnets.

Key Points

  • A bar magnet is a magnetic dipole of moment $m=m_p(2l)$ (units A m$^2$); monopoles do not exist.
  • Dipole field: axial $B=\frac{\mu_0}{4\pi}\frac{2m}{r^3}$, equatorial $B=\frac{\mu_0}{4\pi}\frac{m}{r^3}$; torque $\tau=mB\sin\theta$, energy $U=-mB\cos\theta$.
  • Earth's magnetism: declination ($D$), dip ($I$) and $B_H=B\cos I$, $B_V=B\sin I$, $\tan I=\frac{B_V}{B_H}$.
  • Diamagnetic ($\chi<0$), paramagnetic (small $\chi>0$, Curie's law $\chi\propto1/T$), ferromagnetic (large $\chi>0$).
  • Hysteresis loop area = energy lost per cycle; soft iron (thin loop) for cores, steel (wide loop) for permanent magnets.
Tap an option to check your answer0 / 4
Q1.The SI unit of magnetic dipole moment is:
Explanation: Magnetic moment $m=m_p(2l)$ has units ampere-metre squared (A m$^2$).
Q2.The axial field of a short bar magnet is ___ its equatorial field at the same distance.
Explanation: $B_{axial}=\frac{\mu_0}{4\pi}\frac{2m}{r^3}$ is twice $B_{eq}=\frac{\mu_0}{4\pi}\frac{m}{r^3}$.
Q3.At the magnetic equator the angle of dip is:
Explanation: At the magnetic equator the field is horizontal, so the dip $I=0^\circ$.
Q4.A material with small negative susceptibility is:
Explanation: Diamagnetic materials have $\chi<0$ and $\mu_r$ slightly less than 1.