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#uniform field

24 public questions tagged with this topic.

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 potential energy of a magnetic dipole in a uniform field is highest when:

**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. The potential energy U = -m B cosθ is highest when cosθ = -1 , i.e., θ = 180° , when the dipole is anti-parallel to the field. This is the least stable position, as energy is maximized. Substituting values gives It is anti-parallel to 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 > 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

A magnetic dipole oscillates in a uniform field when displaced from its equilibrium position because:

**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. The torque on a magnetic dipole ( tau = m B sinθ ) acts as a restoring force when displaced from its equilibrium position (aligned with the field). This torque causes oscillatory motion, similar to a pendulum, as it seeks to return to the stable alignment. Substituting values gives The torque acts as a restoring force, 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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Torque on Magnetic Dipole and Potential Energy

A magnetic dipole in a uniform field is in unstable equilibrium when:

**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). The potential energy U = -m B cosθ is maximized when θ = 180° (anti-parallel), making it an unstable equilibrium position. Any small perturbation causes the dipole to rotate toward the stable position ( θ = 0° ), as the energy decreases in that direction. Substituting values gives It is anti-parallel to the field, which matches expected magnitude for this

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

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Electric Dipole - Moment, Field and Torque

What allows an electric dipole to experience a net force in a non-uniform electric field but not in a uniform one?

**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°. In a non-uniform field, the field strength varies across the dipole, causing unequal forces on the positive and negative charges. This results in a net force, unlike in a uniform field where equal and opposite forces cancel out. Substituting values gives Field gradient, 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 Flux

A uniform electric field \( E = 8 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a rectangle

**Flux definition** Φ = ∮ E·dA links field to area orientation. For uniform E perpendicular to surface, Φ = E A, with A = πr² for circle. Inclination reduces flux by cosθ factor, sign indicating outward or inward crossing. Area vector Δ S = 0.2 × 0.3 = 0.06 m² along x-axis. Flux: Φ = E · Δ S = 8 × 10³ × 0.06 = 480 N·m²/C . Substituting values gives 480 N·m²/C, 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 Flux

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Electric Flux

A uniform electric field \( E = 7 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a rectangle

**Measure of field penetration** depends on both magnitude and projected area. Understanding angle between E and normal vector is crucial, flux zero when field parallel to surface, maximum when perpendicular. Area vector Δ S = 0.25 × 0.4 = 0.1 m² along x-axis. Flux: Φ = E · Δ S = 7 × 10³ × 0.1 = 700 N·m²/C . Substituting values gives 700 N·m²/C, 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 Flux

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Electric Flux

A uniform electric field \( E = 6 \times 10^3 \, \text{N/C} \) is along the z-axis. What is the flux through a square of

**Flux definition** Φ = ∮ E·dA links field to area orientation. For uniform E perpendicular to surface, Φ = E A, with A = πr² for circle. Inclination reduces flux by cosθ factor, sign indicating outward or inward crossing. Area vector Δ S = (0.5)² = 0.25 m² along z-axis. Flux: Φ = E · Δ S = 6 × 10³ × 0.25 = 1500 N·m²/C . Substituting values gives 1500 N·m²/C, 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 Flux

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Electric Flux

A uniform electric field \( E = 4 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a square of

**Electric flux** through surface measures field lines crossing it, Φ = E·A = E A cosθ for uniform field, unit N·m²/C. For circular area in xy-plane with field along z, θ = 0°, cosθ = 1, so Φ = E·πR² directly, maximum when field normal to surface. Area vector Δ S = (0.7)² = 0.49 m² along x-axis. Flux: Φ = E · Δ S = 4 × 10³ × 0.49 = 1960 N·m²/C . Substituting values gives 1960 N·m²/C, 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 Flux

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