#uniform magnetic field

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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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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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A magnetic dipole of moment \( 0.8 \, \text{A m}^2 \) is in a uniform field of \( 0.5 \, \text{T} \) at \( 45^\circ \).

**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 = 0.8 A m² , B = 0.5 T , θ = 45° , sin 45° = (1/√(2)) ≈ 0.707 . Substitute: tau = 0.8 × 0.5 × 0.707 ≈ 0.2828 N m ≈ 0.28 N m . Substituting values gives 0.28 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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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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Magnetic Field Due to Bar Magnet - Axial and Equatorial

A magnetic dipole of moment \( 0.2 \, \text{A m}^2 \) is placed in a uniform field of \( 0.3 \, \text{T} \) at \( 45^\ci

**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 = 0.2 A m² , B = 0.3 T , θ = 45° , sin 45° = (1/√(2)) ≈ 0.707 . Substitute: tau = 0.2 × 0.3 × 0.707 = 0.04242 N m ≈ 0.042 N m . Substituting values gives 0.042 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 Lines, Bar Magnet and Dipole Moment

The magnetic field inside a long solenoid is uniform because:

**Bar magnet properties** include dipole moment m = pole strength × separation, unit A·m², field lines emerge from north and enter south outside. Lines never cross, ensuring single valued B at any point, and pattern reflects dipole nature with symmetric loops around magnet. In a long solenoid, the magnetic field is uniform inside due to the symmetrical arrangement of current-carrying loops, which produce overlapping field lines that are parallel and evenly spaced along the solenoid’s axis, minimizing edge effects in the central region. Substituting values gives The field lines are parallel and evenly spaced, which matches expected magnitude for this magnetic configuration, confirming dipole field

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

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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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