Skip to content
Login Register

#magnetic materials

8 public questions tagged with this topic.

Diamagnetism, Paramagnetism and Ferromagnetism

A material with \( B = 0.35 \, \text{T} \) and \( H = 1800 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \tim

**Paramagnetism** has small positive χ ≈ 10⁻³ to 10⁻⁵, weakly attracted towards stronger field, random moments align partially with B, magnetization decreases with temperature following Curie law χ ∝ 1/T. Materials have unpaired electrons with permanent moments. B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.35 T , H = 1800 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.35/4π × 10⁻⁷) ≈ 2.785 × 10⁵ A m⁻¹ . M = 2.785 × 10⁵ - 1800 ≈ 2.767 × 10⁵ A m⁻¹ . Substituting values gives 2.767 × 10⁵ A m⁻¹, which matches

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

View Details
Diamagnetism, Paramagnetism and Ferromagnetism

A material with \( B = 0.4 \, \text{T} \) and \( H = 2000 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \time

**Ferromagnetism** shows large positive χ ≈ 10³ to 10⁵, strong attraction, domain structure with spontaneous magnetization, hysteresis, retentivity. Distinction based on sign and magnitude of χ and behaviour in non-uniform field, explaining attraction versus repulsion. B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.4 T , H = 2000 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.4/4π × 10⁻⁷) ≈ 3.183 × 10⁵ A m⁻¹ . M = 3.183 × 10⁵ - 2000 ≈ 3.163 × 10⁵ A m⁻¹ . Substituting values gives 3.163 × 10⁵ A m⁻¹, which matches expected magnitude

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

View Details
Magnetization, Magnetic Intensity, Susceptibility and Permeability

Which material has a small positive \( \chi \) and aligns weakly with an external magnetic field?

**Relation between B, H, M** is B = μ₀(H+M) = μ₀(1+χ)H. Susceptibility χ = μ_r -1 quantifies material response. Given B, μ_r, n, current I = B/(μ₀ μ_r n), with μ₀ = 4π×10⁻⁷ T·m/A, enabling current calculation for desired B with magnetic core. Paramagnetic materials have a small positive chi and align weakly with an external magnetic field due to the alignment of atomic dipoles. Substituting values gives Paramagnetic, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetization, Magnetic Intensity, Susceptibility and Permeability

View Details
Magnetization, Magnetic Intensity, Susceptibility and Permeability

A material that repels a magnet and is repelled by it in return is likely:

**Magnetization M** is magnetic moment per unit volume (A/m), magnetic intensity H = B/μ₀ - M, susceptibility χ = M/H dimensionless, permeability μ = B/H = μ₀(1+χ), relative permeability μ_r = μ/μ₀ = 1+χ. For solenoid with core, B = μ₀ μ_r n I, n turns per meter (m⁻¹), I current (A). Superconductors, exhibiting perfect diamagnetism (Meissner effect), expel magnetic fields completely, causing mutual repulsion with a magnet. This is distinct from weak diamagnetic repulsion and stronger than other material responses. Substituting values gives Superconductor, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetization, Magnetic Intensity, Susceptibility and Permeability

View Details
Diamagnetism, Paramagnetism and Ferromagnetism

A material that enhances the magnetic field inside it significantly when placed in an external field is likely:

**Ferromagnetism** shows large positive χ ≈ 10³ to 10⁵, strong attraction, domain structure with spontaneous magnetization, hysteresis, retentivity. Distinction based on sign and magnitude of χ and behaviour in non-uniform field, explaining attraction versus repulsion. Ferromagnetic materials greatly enhance the magnetic field inside them due to the strong alignment and growth of magnetic domains, resulting in a high susceptibility and permeability, far exceeding that of paramagnetic or diamagnetic materials. Substituting values gives Ferromagnetic, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

View Details
Diamagnetism, Paramagnetism and Ferromagnetism

A material loses its magnetic properties when heated beyond a certain temperature because:

**Paramagnetism** has small positive χ ≈ 10⁻³ to 10⁻⁵, weakly attracted towards stronger field, random moments align partially with B, magnetization decreases with temperature following Curie law χ ∝ 1/T. Materials have unpaired electrons with permanent moments. Ferromagnetic materials lose their magnetism above the Curie temperature, where thermal energy disrupts the aligned domains, transitioning the material to a paramagnetic state with random moment orientations. Substituting values gives Its domains become disordered, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

View Details
Torque on Magnetic Dipole and Potential Energy

A material has \( B = 0.54 \, \text{T} \) and \( M = 4.0 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \m

**Torque on magnetic dipole** in uniform field B is τ = m × B, magnitude τ = m B sinθ, m moment (A·m²), B field (T), θ angle between m and B (degrees). Torque tends to align moment with field, zero at θ = 0°, maximum mB at 90°, direction given by right-hand rule. B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.54 T , M = 4.0 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.54/4π × 10⁻⁷) ≈ 4.297 × 10⁵ A m⁻¹ . H = 4.297 × 10⁵

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

View Details
Hysteresis, Retentivity, Coercivity and Permanent Magnets

The reason a soft ferromagnetic material loses its magnetism when the external field is removed is:

**Permanent magnet requirement** is high retentivity to maintain field and high coercivity to resist demagnetization. Ability to retain magnetism after field removal is property of hard ferromagnets, related to domain wall pinning and anisotropy. Soft ferromagnetic materials have low coercivity and do not retain significant magnetization after the external field is removed. Their domains, which align strongly in the presence of a field, return to a random orientation due to minimal internal resistance, unlike hard ferromagnets. Substituting values gives Its domains randomize without the field, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Hysteresis, Retentivity, Coercivity and Permanent Magnets

View Details