#dipole

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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Hysteresis, Retentivity, Coercivity and Permanent Magnets

The magnetic potential energy of a dipole with \( m = 0.7 \, \text{A m}^2 \) in a field \( B = 0.2 \, \text{T} \) at \(

**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. U_m = -m B cosθ . Given: m = 0.7 A m² , B = 0.2 T , θ = 0° , cos 0° = 1 . Substitute: U_m = -0.7 × 0.2 × 1 = -0.14 J . Substituting values gives -0.14 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 > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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Diamagnetism, Paramagnetism and Ferromagnetism

A dipole with \( m = 0.7 \, \text{A m}^2 \) in a field \( B = 0.4 \, \text{T} \) at \( 180^\circ \) has potential energy

**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. U_m = -m B cosθ . Given: m = 0.7 A m² , B = 0.4 T , θ = 180° , cos 180° = -1 . U_m = -0.7 × 0.4 × (-1) = 0.28 J . Substituting values gives 0.28 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 > Diamagnetism, Paramagnetism and Ferromagnetism

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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

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Magnetic Field Lines, Bar Magnet and Dipole Moment

Which property of magnetic field lines distinguishes them from electric field lines in the context of a dipole?

**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. Magnetic field lines form continuous closed loops because there are no magnetic monopoles; they emerge from the north pole and enter the south pole, looping back internally. In contrast, electric field lines of a dipole start at the positive charge and end at the negative charge, or extend to infinity if unterminated. Substituting

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

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Electric Dipole - Moment, Field and Torque

A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) is at 60° to a field \( E = 7 \times 10^4 \, \text{N/C} \). What

**Dipole moment** governs torque and energy in external field. Axial field stronger than equatorial, torque maximum at θ = 90°, zero when aligned. Work done rotating dipole relates to ΔU = pE(1 - cosθ), explaining stable equilibrium at θ = 0°. tau = p E sin θ . tau = 6 × 10⁻⁹ × 7 × 10⁴ × sin 60° = 42 × 10⁻⁵ × (√(3)/2) = 3.64 × 10⁻⁴ N m . Substituting values gives 3.64 × 10⁻⁴ N m, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Electric Dipole - Moment, Field and Torque

A dipole with \( p = 9 \times 10^{-9} \, \text{C m} \) is at 90° to a field \( E = 3 \times 10^4 \, \text{N/C} \). What

**Electric dipole** consists of charges +q and -q separated by 2a, dipole moment p = q·2a, vector from negative to positive, unit C·m. In uniform field E, torque τ = p × E, magnitude τ = p E sinθ, tending to align p with E, potential energy U = -p·E = -p E cosθ. tau = p E sin θ . tau = 9 × 10⁻⁹ × 3 × 10⁴ × sin 90° = 27 × 10⁻⁵ × 1 = 2.7 × 10⁻⁴ N m . Substituting values gives 2.7 × 10⁻⁴ N m, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Electric Dipole - Moment, Field and Torque

Why does the electric field due to a dipole have both radial and tangential components at a general point?

**Electric dipole** consists of charges +q and -q separated by 2a, dipole moment p = q·2a, vector from negative to positive, unit C·m. In uniform field E, torque τ = p × E, magnitude τ = p E sinθ, tending to align p with E, potential energy U = -p·E = -p E cosθ. The dipole’s field results from two opposite charges, creating a complex pattern. At a general point, the field vectors from each charge have different directions, resolving into radial (along the line from the dipole) and tangential (perpendicular) components due to asymmetry. Substituting values gives Vector addition, which matches expected magnitude for

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Electric Dipole - Moment, Field and Torque

A dipole with \( p = 8 \times 10^{-9} \, \text{C m} \) is at 90° to a field \( E = 6 \times 10^4 \, \text{N/C} \). What

**Interaction of dipole with uniform field** produces pure couple without net force, equal opposite forces forming torque. Potential energy minimum -pE at alignment, maximum +pE at anti-alignment, governing orientation dynamics. tau = p E sin θ . tau = 8 × 10⁻⁹ × 6 × 10⁴ × sin 90° = 48 × 10⁻⁵ × 1 = 4.8 × 10⁻⁴ N m . Substituting values gives 4.8 × 10⁻⁴ N m, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Electric Dipole - Moment, Field and Torque

A dipole with charges \( +6 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) separated by 2 mm is in a field \( 8 \times 10

**Electric dipole** consists of charges +q and -q separated by 2a, dipole moment p = q·2a, vector from negative to positive, unit C·m. In uniform field E, torque τ = p × E, magnitude τ = p E sinθ, tending to align p with E, potential energy U = -p·E = -p E cosθ. Dipole moment: p = q × 2a = 6 × 10⁻⁶ × 2 × 10⁻³ = 1.2 × 10⁻⁸ C m . Torque: tau = p E sin θ = 1.2 × 10⁻⁸ × 8 × 10⁴ × sin 60° = 9.6 × 10⁻⁴ × (√(3)/2) = 8.31 × 10⁻⁴ N

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Electric Field and Electric Field Lines

What explains why the electric field due to a dipole can be zero at certain points along its equatorial plane?

**Electric field concept** visualizes influence of source charge. Uniform field exerts constant force F = qE, and flux Φ = E·A = E A cosθ links field to area orientation, maximum when field normal to surface. In the equatorial plane, the fields from the dipole’s positive and negative charges are equal in magnitude and opposite in direction at the midpoint. Their vector sum cancels out, resulting in a zero field at that specific location. Substituting values gives Field cancellation, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Field and Electric Field Lines

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Electric Dipole - Moment, Field and Torque

A dipole with \( p = 7 \times 10^{-9} \, \text{C m} \) is at 45° to a field \( E = 4 \times 10^4 \, \text{N/C} \). What

**Dipole moment** governs torque and energy in external field. Axial field stronger than equatorial, torque maximum at θ = 90°, zero when aligned. Work done rotating dipole relates to ΔU = pE(1 - cosθ), explaining stable equilibrium at θ = 0°. tau = p E sin θ . tau = 7 × 10⁻⁹ × 4 × 10⁴ × sin 45° = 28 × 10⁻⁵ × (√(2)/2) = 1.98 × 10⁻⁴ N m . Substituting values gives 1.98 × 10⁻⁴ N m, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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