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#magnetism

118 public questions tagged with this topic.

Why does the absence of magnetic monopoles imply a specific property about magnetic field lines?

**Radio waves** λ≈10⁻¹ to 10⁴ m, f≈10⁴ to 10⁹ Hz, produced by rapid acceleration/deceleration of electrons in aerials/antenna, used for long-distance communication because low frequency diffracts around obstacles and reflects from ionosphere, enabling ground wave and sky wave propagation, effective for broadcasting. The absence of magnetic monopoles, as stated by Gauss’s law for magnetism, implies that magnetic field lines are always closed loops, with no isolated sources or sinks. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Magnetic field

Ref: NCERT > Physics Book > Electromagnetic Waves > Electromagnetic Spectrum - Radio Waves and Microwaves

What does Gauss's Law for magnetism imply in Maxwell's equations?

**Transverse nature** means E and B perpendicular to direction, e.g., wave propagating along z, E along x, B along y, Poynting vector S = E×B/μ₀ along z, energy flow direction. E and B in phase, maxima together, ratio fixed c. Gauss's Law for magnetism, oint B · d A = 0 , implies that there are no magnetic monopoles, as the net magnetic flux through a closed surface is zero. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields No magnetic monopoles exist, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > EM Wave Characteristics - Transverse Nature and E/B Ratio

What is the shape of the magnetic field lines produced by a straight current-carrying wire?

**Magnetic moment of loop** m = N I A (A·m²), potential energy U = -m·B = -N I A B cosθ, torque tends to align m with B. For square side 0.18 m, A = 0.0324 m², N=30, I=2 A, B=0.4 T, θ=60°, τ =30×2×0.0324×0.4×sin60° =0.7776×0.866=0.673 N·m, illustrating large torque for modest parameters. The magnetic field lines around a straight current-carrying wire form concentric circles centered on the wire, as derived from the Biot-Savart law or Ampere’s law. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

A circular loop of radius \( 0.2 \, \text{m} \) with 10 turns carries \( 1.5 \, \text{A} \). What is the magnetic field

**Magnetic field due to long straight wire** at distance r is B = μ₀ I/(2π r), μ₀ = 4π×10⁻⁷ T·m/A, direction circular around wire given by right-hand grip rule. For I = 18 A, r = 0.15 m, B = 2×10⁻⁷×18/0.15 = 2.4×10⁻⁵ T, illustrating 1/r dependence. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 10 × 1.5/2 × 0.2) = (6 π × 10⁻⁶/0.4) = 1.5 π × 10⁻⁵ ≈ 4.71 × 10⁻⁵ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Field Due to Current - Straight Wire and Circular Loop

A material with susceptibility \( \chi = 5 \times 10^{-3} \) has a relative permeability \( \mu_r \) of:

**Geomagnetic field** arises from outer core dynamo, field lines emerge near geographic south pole. Understanding D and I allows conversion between geographic and magnetic coordinates, with B_H = B cos(inclination) used in experiments with tangent galvanometer. μ_r = 1 + chi . Given: chi = 5 × 10⁻³ . Substitute: μ_r = 1 + 5 × 10⁻³ = 1.005 . Substituting values gives 1.005, 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 > Earth's Magnetism and Magnetic Declination

The magnetic potential energy of a dipole with \( m = 0.8 \, \text{A m}^2 \) in a field \( B = 0.25 \, \text{T} \) at \(

**Elements of Earth's field** include declination D, inclination I, horizontal component B_H, total field B = √(B_H² + B_V²). B_H provides compass direction, declination varies with location, important for navigation, inclination 0° at magnetic equator, 90° at poles. U_m = -m B cosθ . Given: m = 0.8 A m² , B = 0.25 T , θ = 180° , cos 180° = -1 . Substitute: U_m = -0.8 × 0.25 × (-1) = 0.2 J . Substituting values gives 0.2 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 > Earth's Magnetism and Magnetic Declination

A dipole with \( m = 0.5 \, \text{A m}^2 \) in a field \( B = 0.6 \, \text{T} \) at \( 180^\circ \) has potential energy

**Earth's magnetism** approximated as dipole inclined to rotation axis, magnetic declination is angle between geographic north and magnetic north, inclination or dip angle is angle between total field and horizontal. Horizontal component B_H = B cosδ, vertical B_V = B sinδ, δ dip angle, B_H ≈ 3-4×10⁻⁵ T in India. U_m = -m B cosθ . Given: m = 0.5 A m² , B = 0.6 T , θ = 180° , cos 180° = -1 . U_m = -0.5 × 0.6 × (-1) = 0.3 J . Substituting values gives 0.3 J, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence

Ref: NCERT > Physics Book > Magnetism and Matter > Earth's Magnetism and Magnetic Declination

The strongest magnetic field of a bar magnet is observed:

**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. The magnetic field of a bar magnet is strongest at its poles, where field lines are most concentrated, as opposed to the central region where the field is weaker and less dense. Substituting values gives At its poles, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m

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

The magnetization \( M \) of a sample is \( 5 \times 10^4 \, \text{A m}^{-1} \) in a magnetic field \( B = 0.1 \, \text{

**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. B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.1 T , M = 5 × 10⁴ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.1/4π × 10⁻⁷) ≈ 7.96 × 10⁴ A m⁻¹ . H = 7.96 × 10⁴ - 5 × 10⁴ = 2.96 × 10⁴ A m⁻¹ ≈ 3 × 10⁴ A m⁻¹ . Substituting values gives

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

The magnetic field contribution \( B_m \) due to a material with \( M = 2 \times 10^5 \, \text{A m}^{-1} \) is: (Take \(

**Hysteresis loop** plots B versus H for ferromagnetic material, retentivity (remanence) is residual B at H=0 after saturation, coercivity is reverse H needed to reduce B to zero. Hard ferromagnets have high retentivity and coercivity, retaining strong magnetism after external field removed, suitable for permanent magnets. B_m = μ₀ M . Given: M = 2 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B_m = 4π × 10⁻⁷ × 2 × 10⁵ = 0.2512 T ≈ 0.25 T . Substituting values gives 0.25 T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B

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

A dipole with \( m = 0.4 \, \text{A m}^2 \) in a field \( B = 0.8 \, \text{T} \) at \( 0^\circ \) has potential energy:

**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 = 0.4 A m² , B = 0.8 T , θ = 0° , cos 0° = 1 . U_m = -0.4 × 0.8 × 1 = -0.32 J . Substituting values gives -0.32 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

A bar magnet with magnetic moment \( 1.5 \, \text{A m}^2 \) is placed at a distance of \( 0.6 \, \text{m} \) along its a

**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 magnetic field along the axis is B = (μ₀/4π) (2m/r³) . Given: m = 1.5 A m² , r = 0.6 m , (μ₀/4π) = 10⁻⁷ . Substitute: B = 10⁻⁷ × (2 × 1.5/(0.6)³) = 10⁻⁷ × (3.0/0.216) ≈ 1.389 × 10⁻⁶ T ≈ 1.39 × 10⁻⁶ T . Substituting values gives 1.39 × 10⁻⁶ T, which matches expected magnitude for this magnetic configuration, confirm

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