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Magnetization, Magnetic Intensity, Susceptibility and Permeability

Latest questions in this category.

30 questions

A material has \( B = 0.28 \, \text{T} \) and \( M = 2.0 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \m

**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. B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.28 T , M = 2.0 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.28/4π × 10⁻⁷) ≈ 2.228 × 10⁵ A m⁻¹ . H = 2.228 × 10⁵ - 2.0 × 10⁵ = 2.28 × 10⁴ A m⁻¹ .

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

The magnetic potential energy of a dipole with \( m = 0.5 \, \text{A m}^2 \) in a field \( B = 0.3 \, \text{T} \) at \(

**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). U_m = -m B cosθ . Given: m = 0.5 A m² , B = 0.3 T , θ = 0° , cos 0° = 1 . Substitute: U_m = -0.5 × 0.3 × 1 = -0.15 J . Substituting values gives -0.15 J, which matches expected magnitude for this magnetic configuration, confirming

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

A material with \( B = 0.66 \, \text{T} \) and \( H = 4500 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \tim

**Magnetic properties** μ_r = 400 indicates 400 times vacuum permeability, so B enhanced 400 times for same nI. H = nI (A/m) for solenoid, M = χ H, B = μ₀(H+M) links microscopic magnetization to macroscopic field. B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.66 T , H = 4500 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.66/4π × 10⁻⁷) ≈ 5.252 × 10⁵ A m⁻¹ . M = 5.252 × 10⁵ - 4500 ≈ 5.207 × 10⁵ A m⁻¹ . Substituting values gives 5.207 × 10⁵ A m⁻¹, which matches

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

A dipole with \( m = 0.2 \, \text{A m}^2 \) in \( B = 0.5 \, \text{T} \) at \( 90^\circ \) has torque:

**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. tau = m B sinθ . Given: m = 0.2 A m² , B = 0.5 T , θ = 90° , sin 90° = 1 . tau = 0.2 × 0.5 × 1 = 0.1 N m . Substituting values gives 0.1 N m, 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

A magnetic dipole in a non-uniform field experiences a net force because:

**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). In a non-uniform field, the field strength varies across the dipole, causing the forces on its poles to differ in magnitude. This imbalance results in a net force, unlike in a uniform field where the forces cancel out. Substituting values gives The field strength varies spatially, which matches expected magnitude for this magnetic

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

The magnetic field contribution \( B_m \) due to a material with \( M = 1.2 \times 10^5 \, \text{A m}^{-1} \) is: (Take

**Magnetic properties** μ_r = 400 indicates 400 times vacuum permeability, so B enhanced 400 times for same nI. H = nI (A/m) for solenoid, M = χ H, B = μ₀(H+M) links microscopic magnetization to macroscopic field. B_m = μ₀ M . Given: M = 1.2 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B_m = 4π × 10⁻⁷ × 1.2 × 10⁵ = 0.15072 T ≈ 0.15 T . Substituting values gives 0.15 T, 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

The magnetic field inside a material becomes zero when:

**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. In a superconductor below its critical temperature, the magnetic field inside becomes zero due to the Meissner effect, where induced currents completely cancel the external field, a hallmark of perfect diamagnetism. Substituting values gives It exhibits perfect diamagnetism, 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

A material with \( B = 0.36 \, \text{T} \) and \( H = 2500 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \tim

**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). B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.36 T , H = 2500 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.36/4π × 10⁻⁷) ≈ 2.864 × 10⁵ A m⁻¹ . M = 2.864 × 10⁵ - 2500 ≈ 2.839

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

The magnetic potential energy of a dipole with \( m = 0.6 \, \text{A m}^2 \) in a field \( B = 0.2 \, \text{T} \) at \(

**Magnetic properties** μ_r = 400 indicates 400 times vacuum permeability, so B enhanced 400 times for same nI. H = nI (A/m) for solenoid, M = χ H, B = μ₀(H+M) links microscopic magnetization to macroscopic field. U_m = -m B cosθ . Given: m = 0.6 A m² , B = 0.2 T , θ = 0° , cos 0° = 1 . Substitute: U_m = -0.6 × 0.2 × 1 = -0.12 J . Substituting values gives -0.12 J, 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

A material with susceptibility \( \chi = 2 \times 10^{-3} \) has a relative permeability \( \mu_r \) of:

**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. μ_r = 1 + chi . Given: chi = 2 × 10⁻³ . Substitute: μ_r = 1 + 2 × 10⁻³ = 1.002 . Substituting values gives 1.002, 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

A dipole with \( m = 0.35 \, \text{A m}^2 \) in \( B = 0.9 \, \text{T} \) at \( 30^\circ \) has torque:

**Magnetic properties** μ_r = 400 indicates 400 times vacuum permeability, so B enhanced 400 times for same nI. H = nI (A/m) for solenoid, M = χ H, B = μ₀(H+M) links microscopic magnetization to macroscopic field. tau = m B sinθ . Given: m = 0.35 A m² , B = 0.9 T , θ = 30° , sin 30° = 0.5 . tau = 0.35 × 0.9 × 0.5 = 0.1575 N m . Substituting values gives 0.1575 N m, 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

A material with \( B = 0.45 \, \text{T} \) and \( H = 3000 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \tim

**Magnetic properties** μ_r = 400 indicates 400 times vacuum permeability, so B enhanced 400 times for same nI. H = nI (A/m) for solenoid, M = χ H, B = μ₀(H+M) links microscopic magnetization to macroscopic field. B = μ₀ (H + M) , so M = (B/μ₀) - H . Given: B = 0.45 T , H = 3000 A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.45/4π × 10⁻⁷) ≈ 3.581 × 10⁵ A m⁻¹ . M = 3.581 × 10⁵ - 3000 ≈ 3.551 × 10⁵ A m⁻¹ . Substituting values gives 3.551 × 10⁵ A m⁻¹, which matches

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