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82 public questions tagged with this topic.

Hysteresis, Retentivity, Coercivity and Permanent Magnets

Unlike a bar magnet, an iron nail near a magnet experiences both a net force and a torque. This is primarily due to:

**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. An iron nail, being a ferromagnetic material, becomes magnetized in the non-uniform field of a bar magnet, developing an induced magnetic moment. The field's gradient causes a net force toward the stronger region (typically the pole), while the misalignment of the induced moment with the field produces a torque. Substituting values gives The non-uniform field inducing a magnetic moment, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence

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

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Diamagnetism, Paramagnetism and Ferromagnetism

A magnetic dipole of moment \( 0.5 \, \text{A m}^2 \) is in a uniform field of \( 0.3 \, \text{T} \) at \( 60^\circ \).

**Diamagnetism** exhibits small negative susceptibility χ ≈ -10⁻⁵ to -10⁻⁶, weakly repelled from stronger to weaker field regions, no permanent moment, induced moment opposite to B, present in all materials but dominated by other effects. Superconductor perfect diamagnet with χ = -1, complete field expulsion. Torque is tau = m B sinθ . Given: m = 0.5 A m² , B = 0.3 T , θ = 60° , sin 60° = (√(3)/2) ≈ 0.866 . Substitute: tau = 0.5 × 0.3 × 0.866 ≈ 0.1299 N m ≈ 0.13 N m . Substituting values gives 0.13 N m, which matches expected magnitude for this

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

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Diamagnetism, Paramagnetism and Ferromagnetism

The absence of a net force on a magnetic dipole in a uniform field implies:

**Diamagnetism** exhibits small negative susceptibility χ ≈ -10⁻⁵ to -10⁻⁶, weakly repelled from stronger to weaker field regions, no permanent moment, induced moment opposite to B, present in all materials but dominated by other effects. Superconductor perfect diamagnet with χ = -1, complete field expulsion. In a uniform magnetic field, the forces on the dipole’s poles are equal in magnitude and opposite in direction, resulting in no net translational force. This occurs because the field strength does not vary, unlike in a non-uniform field where a gradient would produce a net force. Substituting values gives The field strength is constant, which matches expected magnitude for

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

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Diamagnetism, Paramagnetism and Ferromagnetism

A magnetic dipole placed in a uniform magnetic field experiences no net force but a torque. This is because:

**Diamagnetism** exhibits small negative susceptibility χ ≈ -10⁻⁵ to -10⁻⁶, weakly repelled from stronger to weaker field regions, no permanent moment, induced moment opposite to B, present in all materials but dominated by other effects. Superconductor perfect diamagnet with χ = -1, complete field expulsion. In a uniform magnetic field, the forces on the north and south poles of a dipole are equal and opposite, canceling out to produce no net force. However, these forces act at different points, creating a torque that tends to align the dipole with the field. Substituting values gives Forces on the poles cancel out but produce a couple, which

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

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Diamagnetism, Paramagnetism and Ferromagnetism

A magnetic dipole of moment \( 0.15 \, \text{A m}^2 \) is in a uniform field of \( 0.8 \, \text{T} \) at \( 60^\circ \).

**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. Torque is tau = m B sinθ . Given: m = 0.15 A m² , B = 0.8 T , θ = 60° , sin 60° = (√(3)/2) ≈ 0.866 . Substitute: tau = 0.15 × 0.8 × 0.866 ≈ 0.1039 N m ≈ 0.104 N m . Substituting values gives 0.104 N m, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque

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

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Torque on Magnetic Dipole and Potential Energy

When a magnetic dipole is placed perpendicular to a uniform magnetic field, the torque acting on it is maximum because:

**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. The torque on a magnetic dipole is given by tau = m B sinθ . It reaches its maximum value when sinθ = 1 , which occurs at θ = 90° (perpendicular orientation), as the cross product m × B is greatest when the angle between the dipole moment and field

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

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Torque on Magnetic Dipole and Potential Energy

The alignment of a magnetic dipole in a uniform field results in:

**Potential energy of magnetic dipole** U = -m B cosθ explains stability. Given m = 0.9 A·m², B = 0.5 T, θ = 90°, sin90° = 1, τ = 0.45 N·m. For 60°, sin60° = √3/2 ≈0.866, reducing torque proportionally. In a uniform field, a magnetic dipole experiences a torque that aligns it with the field to minimize potential energy ( U = -m B cosθ ), reaching a stable equilibrium when parallel ( θ = 0° ), with no net force due to field uniformity. Substituting values gives A stable equilibrium position, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence

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

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Torque on Magnetic Dipole and Potential Energy

A dipole with \( m = 0.7 \, \text{A m}^2 \) in \( B = 0.2 \, \text{T} \) at \( 45^\circ \) has torque:

**Magnetic dipole in uniform field** experiences torque τ = m B sinθ and potential energy U = -m·B = -m B cosθ, minimum -mB when aligned (θ=0°), maximum +mB at anti-alignment (θ=180°). Work done rotating from θ₁ to θ₂ equals ΔU = mB(cosθ₁ - cosθ₂). tau = m B sinθ . Given: m = 0.7 A m² , B = 0.2 T , θ = 45° , sin 45° = (1/√(2)) ≈ 0.707 . tau = 0.7 × 0.2 × 0.707 = 0.09898 N m ≈ 0.1 N m . Substituting values gives 0.1 N m, which matches expected magnitude for this magnetic configuration, confirming

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

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Magnetic Field Lines, Bar Magnet and Dipole Moment

The reason a bar magnet cannot exert a torque on itself is:

**Magnetic field lines** form continuous closed loops, direction given by tangent at point, density indicates field strength. Unlike electric field lines, magnetic lines never intersect because unique field direction exists at each point, and bar magnet possesses dipole moment m = N I A directed from south to north pole inside magnet. A bar magnet cannot exert a torque on itself because torque requires an external field acting on the dipole. The field produced by the magnet itself does not generate a net rotational effect on its own structure, as internal forces cancel out. Substituting values gives It lacks an external field to act on

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetic Field Lines, Bar Magnet and Dipole Moment

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A magnetic dipole of moment \( 0.3 \, \text{A m}^2 \) is in a uniform field of \( 0.7 \, \text{T} \) at \( 30^\circ \).

**Axial versus equatorial** field comparison shows B_axial = 2 B_eq for same r. Using μ₀ = 4π×10⁻⁷ T·m/A, calculation involves r³ = (0.3)³ = 0.027 m³, so B = 10⁻⁷·m/r³ yields moment estimation. Torque is tau = m B sinθ . Given: m = 0.3 A m² , B = 0.7 T , θ = 30° , sin 30° = 0.5 . Substitute: tau = 0.3 × 0.7 × 0.5 = 0.105 N m . Substituting values gives 0.105 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 > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A magnetic dipole experiences a torque of \( 0.02 \, \text{N m} \) in a field of \( 0.4 \, \text{T} \) at \( 90^\circ \)

**Magnetic field of bar magnet** follows inverse cube law B ∝ m/r³, unlike inverse square for electric dipole. Given B at distance r, moment m = B r³/(μ₀/4π) for equatorial, m = B r³/(2·μ₀/4π) for axial, enabling moment extraction from measured field. tau = m B sinθ , so m = (tau/B sinθ) . Given: tau = 0.02 N m , B = 0.4 T , θ = 90° , sin 90° = 1 . m = (0.02/0.4 × 1) = 0.05 A m² . Substituting values gives 0.05 A m², which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³

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

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Torque on Magnetic Dipole and Potential Energy

A dipole with \( m = 0.9 \, \text{A m}^2 \) in \( B = 0.5 \, \text{T} \) at \( 90^\circ \) has torque:

**Potential energy of magnetic dipole** U = -m B cosθ explains stability. Given m = 0.9 A·m², B = 0.5 T, θ = 90°, sin90° = 1, τ = 0.45 N·m. For 60°, sin60° = √3/2 ≈0.866, reducing torque proportionally. tau = m B sinθ . Given: m = 0.9 A m² , B = 0.5 T , θ = 90° , sin 90° = 1 . tau = 0.9 × 0.5 × 1 = 0.45 N m . Substituting values gives 0.45 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 > Torque on Magnetic Dipole and Potential Energy

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