Q.A closely wound solenoid of $2000$ turns and area of cross-section $1.6 \times 10^{-4}\ \text{m}^2$, carrying a current of $4.0\ \text{A}$, is suspended through its centre allowing it to turn in a horizontal plane.

[!TLDR]

A permanent magnet needs high coercivity, high retentivity and high permeability.

Concept

Magnetic hysteresis (CBSE Class 12 magnetism) tells us that a permanent-magnet material should retain strong magnetism and resist demagnetisation. The relevant properties are retentivity (residual magnetism after removing the field), coercivity (reverse field needed to demagnetise), and permeability (ease of magnetisation).

Solution

For a good permanent magnet:

• High retentivity — so it keeps a strong magnetisation after the magnetising field is removed.
• High coercivity — so it is not easily demagnetised by stray fields, heating or handling.
• High permeability — so it can be magnetised strongly to begin with. …

Concept & Intuition

Paramagnetism arises from unpaired electrons in atoms. Each unpaired electron acts like a tiny bar magnet. In an external magnetic field, these atomic magnets experience a torque that tries to align them with the field. However, thermal agitation fights this alignment, so the net effect is weak and temporary — the material is weakly attracted into the field (unlike ferromagnets, which are strongly attracted). Crucially, the susceptibility $\chi$ is small and positive (typically $10^{-5}$ to $10^{-3}$), meaning the magnetization is in the same direction as the applied field. This is the opposite of diamagnetic materials, which have negative susceptibility and are repelled.

Let’s examine each option:

1. Option A: “Move from a region of strong magnetic field to weak magnetic field”

   This describes diamagnetic behavior. Diamagnets are repelled by magnetic fields, so they seek weaker field regions. Paramagnets are attracted into stronger field regions (because their induced magnetization adds to the field). So A is false for paramagnets.

2. Option B: “Has susceptibility less than zero”

   Susceptibility $\chi < 0$ is the hallmark of diamagnetism. For paramagnetism, $\chi > 0$ (though small). So B is false.

3. Option C: “Attract strongly towards external magnetic field” …

The key concept here is the relationship between magnetic susceptibility, permeability, and the permeability of free space. Susceptibility ($\chi$) tells us how much a material magnetizes in response to an external field. Permeability ($\mu$) measures how easily a magnetic field can form inside the material. They are linked by the simple formula $\mu = \mu_0 (1 + \chi)$. Since the rod is a linear, isotropic material (typical for such problems), this relation is exact.

1. Recall the defining formula The magnetic permeability of a material is given by:

where $\mu_0$ is the permeability of free space and $\chi$ is the magnetic susceptibility. This comes from the fact that the total magnetic field inside the material is the sum of the applied field and the induced magnetization.

2. Plug in the given values We have $\chi = 799$ and $\mu_0 = 4\pi \times 10^{-7} \, \mathrm{T\,m\,A^{-1}}$. So:

3. Simplify the arithmetic

But $32 \times 10^{-5} = 3.2 \times 10^{-4}$, so:

$\mu = 3.2\pi \times 10^{-4} \, \mathrm{T\,m\,A^{-1}}$ …

The key to this question is understanding the physical origin of diamagnetism. Unlike paramagnetism or ferromagnetism, which depend on permanent magnetic moments from unpaired electron spins, diamagnetism is a universal but weak effect caused by the orbital motion of electrons. When an external magnetic field is applied, it induces a change in the orbital motion of electrons (via Lenz’s law), creating a tiny opposing magnetic moment. This is why diamagnetic materials are repelled by magnetic fields.

Let’s examine each option:

1. Option (A): “The material moves from a region of strong magnetic field to weak magnetic field.”

   This is true. Because the induced moment opposes the field, a diamagnetic material experiences a force toward weaker field regions (like a magnet repelling it). This is a classic demonstration—e.g., a diamagnetic bismuth sample is pushed away from a magnet’s pole.

2. Option (B): “The origin of diamagnetism is the spin of electrons.”

   This is false. Electron spin is the source of paramagnetism and ferromagnetism (when spins align). Diamagnetism originates from the orbital angular momentum of electrons—specifically, the induced change in orbital motion. Spin plays no role in pure diamagnetism. This is the incorrect characteristic.

3. Option (C): “Their magnetic susceptibility is small and negative.” …

The key concept here is magnetic susceptibility $\chi$, which measures how much a material becomes magnetized in an external magnetic field. For diamagnetic materials, the induced magnetization is opposite to the field, giving $\chi < 0$. An ideal diamagnetic substance is one that perfectly expels magnetic flux — this is the Meissner effect in superconductors. In such a case, the internal magnetic field is zero, which forces the magnetization $M$ to exactly cancel the applied field $H$, so $M = -H$. Since $\chi = M/H$, we get $\chi = -1$.

Let’s walk through the reasoning step by step:

1. Recall the definition of magnetic susceptibility

   The volume magnetic susceptibility is $\chi = \frac{M}{H}$, where $M$ is magnetization (magnetic moment per unit volume) and $H$ is the applied magnetic field strength. For any material, $\chi$ can be positive (paramagnetic/ferromagnetic) or negative (diamagnetic).

2. Understand diamagnetism

   In ordinary diamagnets (e.g., bismuth, water), the induced magnetic moment is weak and opposite to the field, so $\chi$ is small and negative (typically around $-10^{-5}$). These are not ideal diamagnets.

3. Define “ideal diamagnetic substance”

   In physics, an ideal diamagnet is one that exhibits perfect diamagnetism — it completely expels magnetic flux from its interior. This is the defining property of a superconductor in the Meissner state. Inside such a material, the magnetic flux density $B = 0$.

4. Relate $B$, $H$, and $M$

   The fundamental relation is $B = \mu_0 (H + M)$. For an ideal diamagnet, $B = 0$ inside, so:

5. Compute susceptibility …

With vacuum, $B_0=\mu_0 H$. With a medium of susceptibility $\chi$,

Fractional increase: …

In a paramagnet the atomic magnetic moments tend to align with the applied field, but thermal agitation opposes this alignment. Curie's law quantifies the balance:

$\chi = \frac{C}{T}$ …

The question asks why steel is preferred over soft iron for making permanent magnets. The answer hinges on the magnetic properties that determine how well a material retains its magnetization after the external field is removed.

Concept and Intuition

A permanent magnet needs to hold its magnetic field strongly even when exposed to opposing fields or mechanical shocks. The property that measures this resistance to demagnetization is coercivity — the strength of the reverse magnetic field needed to reduce the material’s magnetization to zero. Soft iron has low coercivity (it magnetizes easily but also demagnetizes easily), while steel (an alloy of iron with carbon) has much higher coercivity due to internal structural defects that pin magnetic domain walls. Thus, steel makes a "hard" magnet that stays magnetized.

Let’s evaluate each option:

1. Option (A): Susceptibility of steel is less than one

   Magnetic susceptibility $\chi$ measures how easily a material magnetizes. For ferromagnetic materials like steel and soft iron, $\chi$ is much greater than 1 (typically hundreds or thousands). While steel’s susceptibility is indeed lower than soft iron’s, this is not the reason for preferring it — in fact, lower susceptibility means it’s harder to magnetize, but that’s a trade-off for stability. The statement is true but irrelevant to the advantage for permanent magnets.

2. Option (B): Permeability of steel is slightly greater than soft iron

   Permeability $\mu = \mu_0(1+\chi)$ is actually higher for soft iron (which has very high permeability, used in electromagnets). Steel has lower permeability. So this statement is false — steel’s permeability is less, not greater.

3. Option (C): Steel has more coercivity than soft iron …

Step 1 — Which law applies?

At $T = 1600$ K, we have $T > T_C$ for both Cobalt ($1400$ K) and Iron ($1000$ K). Above the Curie temperature a ferromagnetic material loses its spontaneous magnetisation and behaves as a paramagnet, whose susceptibility follows the Curie–Weiss law:

where $C$ is the (material) Curie constant.

Step 2 — Write the two susceptibilities

(The problem intends the same Curie constant, so it cancels in the ratio — this is the standard KCET/NCERT treatment.)

Step 3 — Take the ratio …

Concept: relative permeability mu_r = 1 + chi, where chi is the magnetic susceptibility.

• Diamagnetic: chi is small and NEGATIVE, so mu_r slightly less than 1 (mu < mu_0).
• Paramagnetic: chi is small and POSITIVE, so mu_r slightly GREATER than 1 (mu > mu_0). The material is weakly attracted and slightly concentrates the field lines. …

Step 1 — Self inductance.

For a single coil, the flux linkage is proportional to its own current:

So the unit of $L$ is $\dfrac{\text{weber}}{\text{ampere}} = \mathrm{Wb\,A^{-1}} = \text{henry (H)}$.

Step 2 — Mutual inductance.

For two coupled coils, the flux linked with coil 2 due to the current in coil 1 is

Again the unit is $\dfrac{\text{weber}}{\text{ampere}} = \mathrm{Wb\,A^{-1}} = \text{henry}$.

Step 3 — Rule out magnetic flux. …

Concept: relation between relative permeability and magnetic susceptibility.

mu_r = 1 + chi => chi = mu_r - 1.

chi = 6000 - 1 = 5999. …