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Torque on Current Loop, Magnetic Moment and Galvanometer

A wire of length \( 0.8 \, \text{m} \) carrying \( 5 \, \text{A} \) is at \( 30^\circ \) to a magnetic field of \( 0.8 \

**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. Force F = I l B sin θ . F = 5 × 0.8 × 0.8 × sin 30° = 4 × 0.8 × 0.5 = 1.6 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R)

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

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A square loop of side \( 0.16 \, \text{m} \) with 25 turns carries \( 3 \, \text{A} \) in a magnetic field of \( 0.7 \,

**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. Torque tau = N I A B sin θ , where A = 0.16 × 0.16 = 0.0256 m² . tau = 25 × 3 × 0.0256 × 0.7 × sin 45° = 1.344 × 0.707 = 0.9502 ≈ 0.95 N m . Using F = q v B sinθ, F = I l B sinθ, B =

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

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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 1.7 \, \text{m} \) carrying \( 7 \, \text{A} \) is at \( 60^\circ \) to a magnetic field of \( 0.2 \

**Force on current-carrying wire** in magnetic field is F = I l × B, magnitude F = I l B sinθ, I current (A), l length (m), B field (T), θ angle between current direction and B. Direction perpendicular to plane containing wire and B, given by Fleming's left-hand rule. F = I l B sin θ . F = 7 × 1.7 × 0.2 × sin 60° = 2.38 × 0.866 = 2.061 ≈ 2.06 N . 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 > Force on Current-Carrying Conductor and Between Parallel Wires

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A square loop of side \( 0.1 \, \text{m} \) with 25 turns carries \( 2 \, \text{A} \) in a magnetic field of \( 0.8 \, \

**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. Torque tau = N I A B sin θ , where A = 0.1 × 0.1 = 0.01 m² . tau = 25 × 2 × 0.01 × 0.8 × sin 30° = 0.5 × 0.8 × 0.5 = 0.2 N m . Using F = q v B sinθ, F = I l B sinθ, B =

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

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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 2 \, \text{m} \) carrying \( 4 \, \text{A} \) is at \( 60^\circ \) to a magnetic field of \( 0.25 \,

**Force between parallel wires** per unit length is f = μ₀ I₁ I₂/(2π d), μ₀/2π = 2×10⁻⁷ T·m/A, d separation (m), attractive if currents same direction, repulsive if opposite. For I₁=5 A, I₂=7 A, d=0.04 m, f = 2×10⁻⁷×35/0.04 = 1.75×10⁻⁴ N/m, sign indicates repulsion for opposite directions. F = I l B sin θ . F = 4 × 2 × 0.25 × sin 60° = 2 × 0.866 = 1.732 ≈ 1.73 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I A

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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Torque on Current Loop, Magnetic Moment and Galvanometer

In a current-carrying loop placed in a magnetic field, when is the torque maximum?

**SI unit of magnetic field** is tesla (T), defined as force 1 N on 1 A·m wire perpendicular to field. Moving coil galvanometer uses torque τ = N I A B balanced by spring torque k φ, so deflection φ ∝ I, enabling current measurement, with radial field ensuring τ = N I A B always maximum. The torque on a current-carrying loop in a magnetic field is given by tau = N I A B sin θ . It is maximum when sin θ = 1 , i.e., θ = 90° , meaning the plane of the loop is perpendicular to the magnetic field.

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

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Force on Current-Carrying Conductor and Between Parallel Wires

A wire of length \( 1.9 \, \text{m} \) carrying \( 4.5 \, \text{A} \) is at \( 45^\circ \) to a magnetic field of \( 0.3

**Force between parallel wires** per unit length is f = μ₀ I₁ I₂/(2π d), μ₀/2π = 2×10⁻⁷ T·m/A, d separation (m), attractive if currents same direction, repulsive if opposite. For I₁=5 A, I₂=7 A, d=0.04 m, f = 2×10⁻⁷×35/0.04 = 1.75×10⁻⁴ N/m, sign indicates repulsion for opposite directions. Force F = I l B sin θ . F = 4.5 × 1.9 × 0.3 × sin 45° = 8.55 × 0.3 × 0.707 = 1.8127 ≈ 1.81 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Force on Current-Carrying Conductor and Between Parallel Wires

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Earth's Magnetism and Magnetic Declination

A dipole with \( m = 0.2 \, \text{A m}^2 \) in \( B = 0.8 \, \text{T} \) at \( 45^\circ \) has torque:

**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. tau = m B sinθ . Given: m = 0.2 A m² , B = 0.8 T , θ = 45° , sin 45° = (1/√(2)) ≈ 0.707 . tau = 0.2 × 0.8 × 0.707 ≈ 0.11312 N m ≈ 0.11 N m . Substituting values gives 0.11 N m, which matches expected magnitude

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

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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

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Earth's Magnetism and Magnetic Declination

A magnetic dipole of moment \( 0.2 \, \text{A m}^2 \) is in a uniform field of \( 0.9 \, \text{T} \) at \( 30^\circ \).

**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. Torque is tau = m B sinθ . Given: m = 0.2 A m² , B = 0.9 T , θ = 30° , sin 30° = 0.5 . Substitute: tau = 0.2 × 0.9 × 0.5 = 0.09 N m . Substituting values gives 0.09 N m, which matches expected magnitude for this magnetic

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

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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

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Solenoid with Magnetic Core and Magnetic Properties

A dipole with \( m = 0.7 \, \text{A m}^2 \) in \( B = 0.6 \, \text{T} \) at \( 30^\circ \) has torque:

**Core magnetization** M = (μ_r -1)nI, so B = μ₀(nI + M). High μ_r materials like soft iron increase B dramatically for same nI, used in electromagnets, with μ_r up to 5000, enabling strong fields with low current. tau = m B sinθ . Given: m = 0.7 A m² , B = 0.6 T , θ = 30° , sin 30° = 0.5 . tau = 0.7 × 0.6 × 0.5 = 0.21 N m . Substituting values gives 0.21 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 > Solenoid with Magnetic Core and Magnetic Properties

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