Chapter Notes

Magnetism And Matter

20 min read

Introduction

Magnetic phenomena are a universal part of nature, found everywhere from distant galaxies to the tiniest atoms. The Earth itself has a magnetic field that has existed long before humans. The term magnet comes from Magnesia, an island in Greece where magnetic ore was discovered around 600 BC.

We've previously learned that moving charges and electric currents create magnetic fields. In this chapter, we will explore magnetism as a distinct subject.

Here are some fundamental ideas about magnetism:

  • The Earth is a Magnet: Our planet acts like a giant magnet, with its magnetic field pointing roughly from the geographic South Pole to the North Pole.
  • Magnetic Poles: When a bar magnet is suspended freely, it aligns itself in a north-south direction. The end pointing to the geographic north is called the north pole, and the end pointing to the geographic south is the south pole.
  • Attraction and Repulsion: Like poles repel each other (north repels north), while opposite poles attract (north attracts south).
  • No Magnetic Monopoles: You cannot isolate a single north or south pole. If you break a bar magnet in half, you get two new, weaker magnets, each with its own north and south pole. Isolated magnetic poles, known as magnetic monopoles, do not exist, unlike isolated positive and negative electric charges.
  • Making Magnets: It's possible to create magnets using iron and its alloys.

The Bar Magnet

A simple way to visualize a magnet's influence is by sprinkling iron filings on a sheet of glass placed over a bar magnet. The filings arrange themselves in a distinct pattern, revealing the magnet's field. This pattern shows that the magnet has two poles, similar to the positive and negative charges of an electric dipole. One pole is the North pole, and the other is the South pole. A similar pattern is also seen around a current-carrying solenoid.

The Magnetic Field Lines

The pattern formed by iron filings helps us map out magnetic field lines, which are a way to visualize the magnetic field.

Properties of Magnetic Field Lines:

  • They form continuous closed loops. They emerge from the north pole, travel to the south pole, and then continue through the inside of the magnet back to the north pole. This is different from electric field lines, which start on positive charges and end on negative charges.
  • The tangent to a field line at any point gives the direction of the net magnetic field B at that point.
  • The density of the field lines indicates the strength of the magnetic field. Where the lines are closer together, the field is stronger.
  • Magnetic field lines never intersect. If they did, the magnetic field at the point of intersection would have two different directions, which is impossible.
Note
Magnetic field lines are sometimes called magnetic lines of force, but this can be misleading. Unlike in electrostatics, these lines do not show the direction of the force on a moving charge.

Bar Magnet as an Equivalent Solenoid

Ampere's hypothesis suggests that all magnetic phenomena are caused by circulating electric currents. The magnetic field lines of a bar magnet look remarkably similar to those of a current-carrying solenoid. This suggests that a bar magnet can be thought of as being made up of a large number of circulating currents, just like a solenoid.

  • Cutting a Magnet: Cutting a bar magnet in half is like cutting a solenoid. You get two smaller, weaker magnets (or solenoids), and the field lines remain continuous closed loops.
  • Axial Field: The magnetic field along the axis of a solenoid at a large distance (rr) is given by the formula: B=μ04π2mr3B=\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}} This is the same formula for the far axial magnetic field of a bar magnet. This confirms that a bar magnet and a solenoid produce similar magnetic fields, and we can consider a bar magnet to have a magnetic moment equivalent to that of a solenoid producing the same field.

The Dipole in a Uniform Magnetic Field

When a magnetic needle (a small magnet) with a magnetic moment m is placed in a uniform magnetic field B, it experiences a torque.

  • Torque: The torque (τ\tau) on the needle is a turning force that tries to align it with the magnetic field. τ=m×B\tau=\mathbf{m} \times \mathbf{B} The magnitude of the torque is τ=mBsinθ\tau = mB \sin \theta, where θ\theta is the angle between m and B.

  • Potential Energy: The magnetic potential energy (UmU_m) stored in the dipole is given by: Um=mBU_m = -\mathbf{m} \cdot \mathbf{B} This is equal to mBcosθ-mB \cos \theta. The potential energy is defined to be zero when the needle is perpendicular to the field (θ=90\theta = 90^{\circ}).

    • Stable Equilibrium: Potential energy is at its minimum (Um=mBU_m = -mB) when θ=0\theta = 0^{\circ}. This is the most stable position, where the magnetic moment is aligned with the magnetic field.
    • Unstable Equilibrium: Potential energy is at its maximum (Um=+mBU_m = +mB) when θ=180\theta = 180^{\circ}. This is the most unstable position, where the magnetic moment is opposite to the magnetic field.
Example
Conceptual Questions on Bar Magnets

Question (a)

What happens if a bar magnet is cut into two pieces: (i) transverse to its length, (ii) along its length?

Explanation (a)

In both cases, you get two new, complete magnets. Each new piece will have its own north and south pole. Magnetic poles cannot be isolated.


Question (b)

A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why?

Explanation (b)

A uniform magnetic field exerts equal and opposite forces on the two poles of a magnet, resulting in a net force of zero but a net torque. The iron nail, however, is in the non-uniform magnetic field of the bar magnet. This field induces a magnetic moment in the nail. Because the field is non-uniform, the attractive force on the closer pole of the nail is stronger than the repulsive force on the farther pole, resulting in a net attractive force.


Question (c)

Must every magnetic configuration have a north pole and a south pole? What about the field due to a toroid?

Explanation (c)

Not necessarily. This is only true if the source of the field has a net non-zero magnetic moment. A toroid (a donut-shaped coil) or an infinitely long straight wire produces a magnetic field, but they do not have distinct north and south poles. The magnetic field lines form closed loops inside the toroid or around the wire.


Question (d)

You are given two identical-looking iron bars, A and B. One is magnetised, but you don't know which. How can you determine if both are magnetised, or if only one is, which one it is, using only the two bars?

Explanation (d)

  1. Test for Repulsion: Bring one end of bar A close to both ends of bar B. If you find repulsion at any point, both bars are magnetised. Repulsion is the surest test for magnetism, as attraction can occur between a magnet and an unmagnetised piece of iron.
  2. Identify the Magnet: If you only observe attraction, one bar is a magnet and the other is not. To find out which, use the fact that a magnet's field is strongest at its poles (ends) and weakest at its center.
    • Take one bar (say, A) and touch its end to the middle of the other bar (B).
    • If bar A experiences little to no force at the middle of B but is strongly attracted to the ends of B, then B is the magnet.
    • If bar A is attracted with the same force at the middle and ends of B, then A is the magnet.

The Electrostatic Analog

There is a strong analogy between magnetism and electrostatics. The equations for the magnetic field of a bar magnet at large distances are very similar to those for the electric field of an electric dipole. We can get the magnetic equations from the electric ones by making these substitutions:

  • Electric Field E\mathbf{E} \rightarrow Magnetic Field B\mathbf{B}
  • Electric Dipole Moment p\mathbf{p} \rightarrow Magnetic Dipole Moment m\mathbf{m}
  • Electrostatic Constant 14πε0\frac{1}{4 \pi \varepsilon_{0}} \rightarrow Magnetic Constant μ04π\frac{\mu_{0}}{4 \pi}

Using this analogy, we can write the formulas for the magnetic field of a bar magnet (for distances rr much larger than the magnet's length ll):

  • Equatorial Field (BEB_E): The field at a point on the line perpendicular to the magnet's axis. BE=μ0m4πr3\mathbf{B}_{E}=-\frac{\mu_{0} \mathbf{m}}{4 \pi r^{3}}
  • Axial Field (BAB_A): The field at a point along the axis of the magnet. BA=μ04π2mr3\mathbf{B}_{A}=\frac{\mu_{0}}{4 \pi} \frac{2 \mathbf{m}}{r^{3}}

The following table summarizes the analogy:

ElectrostaticsMagnetism
Constant1/ε01 / \varepsilon_{0}μ0\mu_{0}
Dipole momentpm
Equatorial Fieldp/4πε0r3-\mathrm{p} / 4 \pi \varepsilon_{0} r^{3}μ0m/4πr3-\mu_{0} \mathbf{m} / 4 \pi r^{3}
Axial Field2p/4πε0r32 \mathrm{p} / 4 \pi \varepsilon_{0} r^{3}μ02m/4πr3\mu_{0} 2 \mathbf{m} / 4 \pi r^{3}
Torque in External Fieldp×E\mathbf{p} \times \mathbf{E}m×B\mathbf{m} \times \mathbf{B}
Energy in External Field-pE-mB

Magnetism and Gauss's Law

In electrostatics, Gauss's Law states that the net electric flux through a closed surface is proportional to the net charge enclosed. If a closed surface encloses a positive charge, there is a net outward flux of electric field lines.

The situation for magnetism is fundamentally different. Since there are no isolated magnetic poles (monopoles), magnetic field lines always form closed loops. They don't start or end anywhere.

Gauss's Law for Magnetism states that the net magnetic flux through any closed surface is always zero. ϕB=allBΔS=0\phi_{B}=\sum_{all} \mathbf{B} \cdot \Delta \mathbf{S}=0

This means that for any closed surface, the number of magnetic field lines entering the surface is exactly equal to the number of lines leaving it. This law is a mathematical expression of the fact that magnetic monopoles do not exist. The simplest magnetic element is a dipole or a current loop.

Example
If magnetic monopoles existed, how would Gauss's law of magnetism be modified?

Explanation

If magnetic monopoles (with magnetic charge qmq_m) existed, Gauss's law for magnetism would look very similar to Gauss's law for electrostatics. The right-hand side of the equation would no longer be zero but would be proportional to the net magnetic charge enclosed by the surface: SBΔs=μ0qm\int_{\mathrm{S}} \mathbf{B} \cdot \Delta \mathbf{s}=\mu_{0} q_{m}

Magnetisation and Magnetic Intensity

When a material is placed in an external magnetic field, it can become magnetised. To describe this, we use a few important quantities.

  • Magnetisation (M): This is the net magnetic moment per unit volume of a material. It's a measure of how magnetised a substance has become. M=mnetV\mathbf{M}=\frac{\mathbf{m}_{net}}{V} The unit of magnetisation is Amperes per meter (A m1\text{A m}^{-1}).

When a material is placed inside a current-carrying solenoid, the total magnetic field B inside the material is the sum of the original field from the solenoid (B0\mathbf{B}_0) and the field contributed by the magnetisation of the material (Bm\mathbf{B}_m). B=B0+Bm\mathbf{B}=\mathbf{B}_{0}+\mathbf{B}_{\mathrm{m}} The field due to the material's magnetisation is given by Bm=μ0M\mathbf{B}_{\mathrm{m}}=\mu_{0} \mathbf{M}.

  • Magnetic Intensity (H): This is a vector field that represents the part of the total magnetic field that comes from external sources, like the current in a solenoid. It is defined as: H=Bμ0M\mathbf{H}=\frac{\mathbf{B}}{\mu_{0}}-\mathbf{M} The total magnetic field can then be written as B=μ0(H+M)\mathbf{B}=\mu_{0}(\mathbf{H}+\mathbf{M}). The unit of H is also A m1\text{A m}^{-1}.

  • Magnetic Susceptibility (χ\chi): For many materials, the magnetisation M is directly proportional to the magnetic intensity H. The constant of proportionality is the magnetic susceptibility, χ\chi. M=χH\mathbf{M}=\chi \mathbf{H} χ\chi is a dimensionless quantity that measures how easily a material can be magnetised.

    • For paramagnetic materials, χ\chi is small and positive.
    • For diamagnetic materials, χ\chi is small and negative.
  • Magnetic Permeability (μ\mu): We can relate the total magnetic field B directly to the magnetic intensity H. B=μ0(1+χ)H=μH\mathbf{B}=\mu_{0}(1+\chi) \mathbf{H} = \mu \mathbf{H} Here, μ=μ0(1+χ)\mu = \mu_0(1+\chi) is the magnetic permeability of the material.

  • Relative Magnetic Permeability (μr\mu_r): This is a dimensionless quantity defined as: μr=1+χ\mu_{r}=1+\chi It relates the permeability of the material to the permeability of free space: μ=μ0μr\mu = \mu_0 \mu_r.

Example
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2 A. If the number of turns is 1000 per metre, calculate (a) H, (b) B, (c) M and (d) the magnetising current ImI_m.

Given

  • Relative permeability, μr=400\mu_r = 400
  • Current, I=2.0 AI = 2.0 \text{ A}
  • Number of turns per metre, n=1000 m1n = 1000 \text{ m}^{-1}
  • Permeability of free space, μ0=4π×107 N/A2\mu_0 = 4\pi \times 10^{-7} \text{ N/A}^2

To Find

(a) Magnetic intensity, HH (b) Magnetic field, BB (c) Magnetisation, MM (d) Magnetising current, ImI_m

Formula

H=nIH = nI B=μrμ0HB = \mu_r \mu_0 H M=(μr1)HM = (\mu_r - 1)H The magnetising current is the additional current needed to produce the same B field without the core.

Solution

(a) Calculate Magnetic Intensity (H) The magnetic intensity H depends only on the external current, not the core material. H=nI=(1000 m1)×(2.0 A)=2×103 A/mH = nI = (1000 \text{ m}^{-1}) \times (2.0 \text{ A}) = 2 \times 10^3 \text{ A/m} Answer for part (a) = 2×103 A/m2 \times 10^3 \text{ A/m}


(b) Calculate Magnetic Field (B) The total magnetic field B inside the core is enhanced by the material. B=μrμ0H=400×(4π×107 N/A2)×(2×103 A/m)B = \mu_r \mu_0 H = 400 \times (4\pi \times 10^{-7} \text{ N/A}^2) \times (2 \times 10^3 \text{ A/m}) B1.0 TB \approx 1.0 \text{ T} Answer for part (b) = 1.0 T1.0 \text{ T}


(c) Calculate Magnetisation (M) Magnetisation is the contribution of the material to the magnetic field. M=(μr1)H=(4001)×(2×103 A/m)=399×2×103 A/mM = (\mu_r - 1)H = (400 - 1) \times (2 \times 10^3 \text{ A/m}) = 399 \times 2 \times 10^3 \text{ A/m} M8×105 A/mM \approx 8 \times 10^5 \text{ A/m} Answer for part (c) = 8×105 A/m8 \times 10^5 \text{ A/m}


(d) Calculate Magnetising Current (ImI_m) The magnetising current is the equivalent current that would produce the field contribution from the material. The total field B is equivalent to the field from a solenoid with current (I+IM)(I + I_M) and no core. The field due to the material is Bm=BB0=Bμ0HB_m = B - B_0 = B - \mu_0 H. This field is also given by Bm=μ0nIMB_m = \mu_0 n I_M. Therefore, μ0nIM=Bμ0H\mu_0 n I_M = B - \mu_0 H. IM=Bμ0Hμ0n=1.0 T(4π×107 N/A2)(2×103 A/m)(4π×107 N/A2)(1000 m1)794 AI_M = \frac{B - \mu_0 H}{\mu_0 n} = \frac{1.0 \text{ T} - (4\pi \times 10^{-7} \text{ N/A}^2)(2 \times 10^3 \text{ A/m})}{(4\pi \times 10^{-7} \text{ N/A}^2)(1000 \text{ m}^{-1})} \approx 794 \text{ A} Alternatively, we can find M=nIMM = nI_M, so IM=M/n=(8×105 A/m)/1000 m1=800 AI_M = M/n = (8 \times 10^5 \text{ A/m}) / 1000 \text{ m}^{-1} = 800 \text{ A}. The slight difference is due to approximation. The text gets 794 A, let's follow its logic: M=(μr1)H=399×2000=7.98×105 A/mM = (\mu_r-1)H = 399 \times 2000 = 7.98 \times 10^5 \text{ A/m}. The magnetising current can be thought of as IM=M/n=7.98×105/1000=798 AI_M = M/n = 7.98 \times 10^5 / 1000 = 798 \text{ A}. A more direct way is M=(μr1)H=(μr1)nIM = (\mu_r-1)H = (\mu_r-1)nI, and the current that produces this magnetization is IM=(μr1)I=399×2=798 AI_M = (\mu_r-1)I = 399 \times 2 = 798 \text{ A}. The book's calculation seems to use a different approach. Let's re-examine the source text method: Thus B = mu_r n(I+I_M). This is an error in the source text. The formula should be B = mu_0 n(I+I_M). Let's re-calculate using the correct formula B = mu_0 n(I+I_M): 1.0 T=(4π×107)(1000)(2+IM)1.0 \text{ T} = (4\pi \times 10^{-7})(1000)(2 + I_M) 1.04π×104=2+IM\frac{1.0}{4\pi \times 10^{-4}} = 2 + I_M 795.77=2+IM    IM794 A795.77 = 2 + I_M \implies I_M \approx 794 \text{ A} This matches the book's answer. The magnetising current is the additional current that would be needed in an air-core solenoid to produce the same total field B.

Answer for part (d) = 794 A794 \text{ A}

Magnetic Properties of Materials

Based on their response to an external magnetic field, materials can be classified into three main types: diamagnetic, paramagnetic, and ferromagnetic.

DiamagneticParamagneticFerromagnetic
Susceptibility (χ\chi)1χ<0-1 \leq \chi < 00<χ<ε0 < \chi < \varepsilon (small positive)χ1\chi \gg 1 (large positive)
Relative Permeability (μr\mu_r)0μr<10 \leq \mu_r < 11<μr<1+ε1 < \mu_r < 1+\varepsilonμr1\mu_r \gg 1
Permeability (μ\mu)μ<μ0\mu < \mu_0μ>μ0\mu > \mu_0μμ0\mu \gg \mu_0

Diamagnetism

Diamagnetic substances are those that are weakly repelled by a magnetic field. They tend to move from a stronger part of a magnetic field to a weaker part.

  • Behavior: When a diamagnetic material is placed in a magnetic field, the field lines are repelled or expelled from it, and the magnetic field inside the material is slightly reduced.
  • Cause: In diamagnetic materials, the atoms have no net magnetic moment. When an external field is applied, it slightly changes the motion of the electrons orbiting the nucleus. This induces a small magnetic moment in the opposite direction to the applied field, causing repulsion. This effect is present in all substances but is often masked by stronger paramagnetic or ferromagnetic effects.
  • Examples: Bismuth, copper, lead, silicon, water, and sodium chloride.
  • Superconductors: These are the most extreme diamagnetic materials. When cooled to very low temperatures, they become perfect diamagnets, completely expelling all magnetic field lines from their interior. For a superconductor, χ=1\chi = -1 and μr=0\mu_r = 0. This phenomenon is called the Meissner effect.

Paramagnetism

Paramagnetic substances are those that are weakly attracted to a magnetic field. They tend to move from a weaker part of a magnetic field to a stronger part.

  • Behavior: When a paramagnetic material is placed in a magnetic field, the field lines become slightly more concentrated inside it, and the field is enhanced.
  • Cause: The individual atoms of paramagnetic materials have a permanent magnetic dipole moment. However, due to random thermal motion, these moments are randomly oriented, resulting in no net magnetisation. When an external field is applied, it tends to align these atomic dipoles in the direction of the field, creating a weak net magnetisation and thus a weak attraction.
  • Examples: Aluminium, sodium, calcium, oxygen, and copper chloride.

Ferromagnetism

Ferromagnetic substances are those that are strongly attracted to a magnetic field. They show a strong tendency to move from a weak field region to a strong field region.

  • Behavior: In a ferromagnetic material, the magnetic field lines are highly concentrated, and the field inside is strongly enhanced.
  • Cause: Like paramagnetic materials, ferromagnetic atoms have permanent magnetic moments. However, a quantum mechanical interaction causes these moments to spontaneously align with their neighbors in large regions called domains. Each domain is fully magnetised.
    • In an unmagnetised material, the domains are randomly oriented, so there is no net magnetisation.
    • When an external field is applied, domains aligned with the field grow in size, and other domains rotate to align with the field. This results in a very strong net magnetisation.
  • Hard and Soft Ferromagnets:
    • Hard Ferromagnets: Materials like Alnico that retain their magnetisation even after the external field is removed. They are used to make permanent magnets.
    • Soft Ferromagnets: Materials like soft iron that lose their magnetisation when the external field is removed. They are used in electromagnets and transformer cores.
  • Effect of Temperature: Ferromagnetism is temperature-dependent. Above a certain temperature (the Curie temperature), a ferromagnetic material loses its domain structure and becomes paramagnetic.
  • Examples: Iron, cobalt, nickel, gadolinium.

Congratulations! You've completed this chapter

Great job reading through all sections. Ready to test your knowledge and reinforce your learning?