#magnetic torque

Hysteresis, Retentivity, Coercivity and Permanent Magnets

A dipole with \( m = 0.5 \, \text{A m}^2 \) in \( B = 0.1 \, \text{T} \) at \( 60^\circ \) has torque:

**Soft ferromagnetic materials** have low coercivity and retentivity, narrow hysteresis loop, lose magnetism when external field removed, ideal for electromagnets and transformer cores. Energy loss per cycle proportional to loop area, explaining why soft materials minimize loss. tau = m B sinθ . Given: m = 0.5 A m² , B = 0.1 T , θ = 60° , sin 60° = (√(3)/2) ≈ 0.866 . tau = 0.5 × 0.1 × 0.866 = 0.0433 N m ≈ 0.043 N m . Substituting values gives 0.043 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 > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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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

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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

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A magnetic dipole of moment \( 0.25 \, \text{A m}^2 \) is in a uniform field of \( 0.5 \, \text{T} \) at \( 30^\circ \).

**Field due to bar magnet** on axial line is B_axial = (μ₀/4π)·2m/r³, equatorial B_eq = (μ₀/4π)·m/r³, where μ₀/4π = 10⁻⁷ T·m/A, m magnetic moment (A·m²), r distance (m). Axial field twice equatorial at same distance and parallel to moment, equatorial opposite to moment direction. Torque is tau = m B sinθ . Given: m = 0.25 A m² , B = 0.5 T , θ = 30° , sin 30° = 0.5 . Substitute: tau = 0.25 × 0.5 × 0.5 = 0.0625 N m . Substituting values gives 0.0625 N m, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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