#uniform magnetic field

Magnetic Force on Moving Charge - Lorentz Force and Motion

What is the direction of the magnetic force on a positive charge moving parallel to a uniform magnetic field?

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. The magnetic force is given by F = q (v × B) . If the velocity v is parallel to the magnetic field B , the angle between them is 0° , so sin 0° = 0 , and the force magnitude is zero. Thus, there is no force. Using F = q v B sinθ, F

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

View Details

Hysteresis, Retentivity, Coercivity and Permanent Magnets

The torque on a magnetic dipole in a uniform magnetic field is zero when:

**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. The torque on a magnetic dipole is given by tau = m B sinθ . It becomes zero when sinθ = 0 , which occurs when the dipole is aligned with the field ( θ = 0° ) or anti-aligned ( θ = 180° ), as the cross product m × B vanishes. Substituting values gives The dipole is parallel or anti-parallel to the field, which matches

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

View Details

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. Substitu

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

View Details

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

View Details

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

View Details

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

View Details

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

View Details

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

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

View Details

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

View Details