Magnetism And Matter: CBSE Class 12 Physics

Magnetic Field and Dipole Moment

Just as an electric charge creates an electric field, a moving charge or a current loop produces a magnetic field. We can visualise magnetic fields using magnetic field lines, which emerge from the North Pole and enter the South Pole outside the magnet, forming continuous closed loops. The density of these lines indicates the strength of the field. A bar magnet is an elementary magnetic dipole. Its strength and orientation are described by its magnetic dipole moment (\( \vec{M} \)), which points from the South Pole to the North Pole. When placed in an external uniform magnetic field (\( \vec{B} \)), a magnetic dipole experiences a torque (\( \vec{\tau} = \vec{M} \times \vec{B} \)) that tends to align it with the field. The potential energy of a magnetic dipole in a magnetic field is given by \( U = -\vec{M} \cdot \vec{B} \).

An interesting equivalence exists between a bar magnet and a current-carrying solenoid. A solenoid behaves like a bar magnet, with its magnetic dipole moment given by \( M = nIA \), where \( n \) is the number of turns per unit length, \( I \) is the current, and \( A \) is the cross-sectional area. This equivalence is crucial for understanding how magnetic fields are generated and interact. The Earth itself acts like a giant bar magnet, producing its own magnetic field. This field is not uniform and changes from place to place. The study of Earth's magnetism involves understanding magnetic declination (the angle between geographic and magnetic meridians) and magnetic dip (the angle a compass needle makes with the horizontal at a location), along with the horizontal component of Earth's magnetic field.

Key Terms in Magnetism

Magnetic Intensity (H)

The degree to which a magnetic field can magnetise a material. It is related to the external magnetising field and is measured in Ampere per metre (A/m).

Magnetisation (M)

The net magnetic dipole moment per unit volume induced in a material when placed in an external magnetic field. It represents how strongly a material is magnetised and is measured in Ampere per metre (A/m).

Magnetic Permeability (μ)

A measure of a material's ability to support the formation of a magnetic field within itself. It is the ratio of magnetic induction (B) to magnetic intensity (H), i.e., \( \mu = B/H \). Its unit is Tesla metre per Ampere (T m/A) or Henry per metre (H/m).

Relative Permeability (μr)

The ratio of the permeability of a material (μ) to the permeability of free space (μ₀). It indicates how much more effectively a given material can concentrate magnetic flux than a vacuum. \( \mu_r = \mu / \mu_0 \).

Magnetic Susceptibility (χm)

A dimensionless quantity that indicates the degree to which a material can be magnetised in response to an applied magnetic field. It is the ratio of magnetisation (M) to magnetic intensity (H), i.e., \( \chi_m = M/H \). It is related to relative permeability by \( \mu_r = 1 + \chi_m \).

Classification of Magnetic Materials

Materials respond differently when placed in an external magnetic field, primarily due to the magnetic moments of their constituent atoms and how these moments align. This difference in response allows us to classify materials into three main categories:

1. Diamagnetic Materials: These materials are weakly repelled by a magnetic field. They have no permanent atomic magnetic dipole moments. When an external magnetic field is applied, it induces a weak magnetic moment in the opposite direction to the applied field, hence they move from stronger to weaker parts of the field. Their magnetic susceptibility (\( \chi_m \)) is small and negative (e.g., -10⁻⁵). Their relative permeability (\( \mu_r \)) is slightly less than 1. Examples include water, copper, bismuth, silicon, nitrogen, and sodium chloride.

2. Paramagnetic Materials: These materials are weakly attracted to a magnetic field. Their atoms possess permanent magnetic dipole moments, but in the absence of an external field, these moments are randomly oriented due to thermal agitation, resulting in zero net magnetisation. When an external magnetic field is applied, these moments partially align with the field, causing a weak net magnetisation in the direction of the field. They move from weaker to stronger parts of the field. Their magnetic susceptibility (\( \chi_m \)) is small and positive (e.g., 10⁻³ to 10⁻⁵). Their relative permeability (\( \mu_r \)) is slightly greater than 1. The magnetisation of paramagnetic materials is inversely proportional to the absolute temperature (Curie's Law). Examples include aluminium, sodium, calcium, oxygen, platinum, and copper chloride.

3. Ferromagnetic Materials: These materials are strongly attracted to a magnetic field and can retain their magnetisation even after the external field is removed (hysteresis). They possess permanent, large atomic magnetic moments that align spontaneously in regions called 'domains'. Within a domain, all atomic moments are aligned, but domains themselves are randomly oriented, resulting in zero net magnetisation in the absence of an external field. When an external magnetic field is applied, domains aligned with the field grow at the expense of others, and misaligned domains rotate to align with the field, leading to a very strong magnetisation. Their magnetic susceptibility (\( \chi_m \)) is very large and positive (e.g., hundreds or thousands). Their relative permeability (\( \mu_r \)) is much greater than 1. They exhibit complex behaviour like hysteresis and can be demagnetised above a certain temperature called the Curie temperature. Examples include iron, cobalt, nickel, and their alloys.

Worked Examples

• Example 1: Torque on a Bar Magnet A short bar magnet of magnetic moment \( 0.4 \text{ J T}^{-1} \) is placed in a uniform magnetic field of \( 0.16 \text{ T} \). The magnet is in stable equilibrium when its magnetic moment is parallel to the field. What is the work done in turning the magnet from stable equilibrium to an orientation where its magnetic moment is antiparallel to the field? Step 1: Understand the initial and final states. In stable equilibrium, the magnetic moment (\( \vec{M} \)) is parallel to the magnetic field (\( \vec{B} \)), so the angle \( \theta_1 = 0^\circ \). The potential energy is \( U_1 = -MB\cos(0^\circ) = -MB \). When the magnetic moment is antiparallel to the field, \( \theta_2 = 180^\circ \). The potential energy is \( U_2 = -MB\cos(180^\circ) = -MB(-1) = +MB \). Step 2: Calculate the work done. The work done in turning the magnet is equal to the change in potential energy: \( W = U_2 - U_1 = (+MB) - (-MB) = 2MB \) Given \( M = 0.4 \text{ J T}^{-1} \) and \( B = 0.16 \text{ T} \). \( W = 2 \times 0.4 \text{ J T}^{-1} \times 0.16 \text{ T} = 0.8 \times 0.16 \text{ J} = 0.128 \text{ J} \). Final Answer: The work done is \( 0.128 \text{ J} \).
• Example 2: Magnetic Field at Axial Point of a Bar Magnet A short bar magnet has a magnetic moment of \( 5.0 \text{ A m}^2 \). Calculate the magnitude of the magnetic field (induction) at a distance of \( 0.1 \text{ m} \) from its centre on its axial line. (Given: \( \mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1} \)). Step 1: Recall the formula for magnetic field on axial line. For a short bar magnet, the magnetic field (B) at an axial point at distance \( r \) from its centre is given by: \( B = \frac{\mu_0}{4\pi} \frac{2M}{r^3} \) Step 2: Substitute the given values into the formula. Given \( M = 5.0 \text{ A m}^2 \), \( r = 0.1 \text{ m} \), and \( \frac{\mu_0}{4\pi} = 10^{-7} \text{ T m A}^{-1} \). \( B = (10^{-7} \text{ T m A}^{-1}) \times \frac{2 \times 5.0 \text{ A m}^2}{(0.1 \text{ m})^3} \) \( B = 10^{-7} \times \frac{10}{0.001} \text{ T} \) \( B = 10^{-7} \times \frac{10}{10^{-3}} \text{ T} \) \( B = 10^{-7} \times 10 \times 10^3 \text{ T} \) \( B = 10^{-7} \times 10^4 \text{ T} = 10^{-3} \text{ T} \). Final Answer: The magnitude of the magnetic field at the axial point is \( 10^{-3} \text{ T} \).

Exam Tips for Magnetism and Matter

When tackling numerical problems, always pay close attention to the units and ensure consistency. Remember that magnetic moment (\( M \)) is a vector quantity, and its direction is crucial in torque and potential energy calculations. For Earth's magnetism, distinguish carefully between geographic and magnetic meridians when defining declination. For magnetic materials, thoroughly understand the differences in their properties (magnetic susceptibility \( \chi_m \), relative permeability \( \mu_r \), and their temperature dependence like Curie's Law for paramagnets and Curie temperature for ferromagnets). Many conceptual questions revolve around these distinctions. Practise deriving expressions for magnetic fields at axial and equatorial points of a short bar magnet.

Practice Questions with Solutions

• Q: A bar magnet of magnetic moment \( M \) is cut into two equal pieces along its length. What is the magnetic moment of each piece? What if it is cut perpendicular to its length? A: Step 1: When cut along its length, each piece retains the original pole strength but its length is halved. The magnetic moment is \( m \times (l/2) \), where \( m \) is pole strength and \( l \) is length. Since the area perpendicular to the length is also halved, effectively \( M' = M/2 \). Step 2: When cut perpendicular to its length, each piece has half the original length (l/2) but also has a reduced pole strength (since the poles are now closer to the center of the original magnet, effectively). However, considering it as two new magnets, each piece will have its own new North and South poles. The pole strength remains the same, but the effective magnetic length becomes \( l/2 \). So, the magnetic moment of each piece becomes \( m \times (l/2) = M/2 \). Final answer: In both cases, the magnetic moment of each piece is \( M/2 \).
• Q: What are magnetic field lines? List two properties of magnetic field lines. A: Step 1: Magnetic field lines are imaginary lines used to represent the direction and strength of a magnetic field at various points. They are the path along which a hypothetical isolated North Pole would tend to move. Step 2: Two properties are: (1) They form continuous closed loops, extending from the North pole to the South pole outside the magnet and from the South pole to the North pole inside the magnet. (2) No two magnetic field lines ever intersect each other. If they did, it would mean that at the point of intersection, the compass needle would point in two directions, which is impossible. Final answer: Magnetic field lines are imaginary lines indicating magnetic field direction and strength. They form continuous closed loops and never intersect.
• Q: Differentiate between diamagnetic and paramagnetic materials based on their magnetic susceptibility and behaviour in an external magnetic field. A: Step 1: Magnetic Susceptibility (\( \chi_m \)): For diamagnetic materials, \( \chi_m \) is small and negative. For paramagnetic materials, \( \chi_m \) is small and positive. Step 2: Behaviour in an external magnetic field: Diamagnetic materials are weakly repelled by an external magnetic field and tend to move from stronger to weaker regions of the field. Paramagnetic materials are weakly attracted to an external magnetic field and tend to move from weaker to stronger regions of the field. Final answer: Diamagnetic materials have small, negative susceptibility and are repelled by magnetic fields, while paramagnetic materials have small, positive susceptibility and are weakly attracted.
• Q: A long solenoid has 2000 turns per metre and carries a current of \( 2.5 \text{ A} \). A short bar magnet of magnetic moment \( 1.5 \text{ J T}^{-1} \) is placed inside the solenoid with its axis parallel to the solenoid's axis. What is the force and torque on the bar magnet? A: Step 1: Calculate the magnetic field inside the solenoid. The magnetic field \( B \) inside a long solenoid is given by \( B = \mu_0 n I \), where \( n \) is turns per unit length and \( I \) is current. \( B = (4\pi \times 10^{-7} \text{ T m A}^{-1}) \times (2000 \text{ m}^{-1}) \times (2.5 \text{ A}) \) \( B = (4\pi \times 10^{-7}) \times 5000 \text{ T} = 20000\pi \times 10^{-7} \text{ T} \approx 0.00628 \text{ T} \). Step 2: Determine the force on the bar magnet. Since the magnetic field inside an ideal long solenoid is uniform, a magnetic dipole (bar magnet) placed in a uniform magnetic field experiences no net translational force. Step 3: Determine the torque on the bar magnet. The magnet's axis is parallel to the solenoid's axis, which means the magnetic moment \( \vec{M} \) is parallel to the magnetic field \( \vec{B} \). The angle \( \theta \) between \( \vec{M} \) and \( \vec{B} \) is \( 0^\circ \). Torque \( \vec{\tau} = \vec{M} \times \vec{B} = MB\sin\theta \). Since \( \sin(0^\circ) = 0 \), the torque is zero. Final answer: The force on the bar magnet is zero, and the torque on the bar magnet is also zero.

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