#torque

Conductors, Electrostatic Shielding and Dielectrics

A dipole \( p = 9 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 180^\circ \) in a field

**Conductor in electrostatic equilibrium** has E=0 inside, charges reside on surface, potential constant throughout conductor. Hollow shell with no internal charge has zero field inside cavity, even if external field present, charges on outer surface screen interior, principle used in Faraday cage. Work done: W = p E (cos θ₀ - cos θ₁) = 9 × 10⁻⁹ × 5 × 10⁴ × (cos 90° - cos 180°) . W = 9 × 10⁻⁹ × 5 × 10⁴ × (0 - (-1)) = 9 × 10⁻⁹ × 5 × 10⁴ × 1 = 4.5 × 10⁻⁴ J . Using V = kQ/r, U = k q₁q₂/r,

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Conductors, Electrostatic Shielding and Dielectrics

A dipole \( p = 10 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 90^\circ \) in a field \

**Electrostatic shielding** inside hollow conducting shell field zero when no charges inside, regardless of external field, because free charges redistribute on outer surface to cancel external field inside conductor, E=0 inside material in equilibrium, consequence of Gauss's law and conductor property. Work done: W = p E (cos θ₀ - cos θ₁) = 10 × 10⁻⁹ × 4 × 10⁵ × (cos 0° - cos 90°) . W = 10 × 10⁻⁹ × 4 × 10⁵ × (1 - 0) = 4 × 10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Equipotential Surfaces and Relation Between Field and Potential

A dipole \( p = 4 \times 10^{-10} \, \text{C m} \) is in a uniform field \( E = 10^5 \, \text{N/C} \) at \( 60^\circ \).

**Equipotential through midpoint** of +q and -q is perpendicular bisector, V=0 everywhere on it because contributions k q/r and k(-q)/r cancel. For two opposite charges, potential at midpoint zero, but field non-zero, pointing from positive to negative, illustrating vector vs scalar nature. U = -p · E = -p E cos θ = -4 × 10⁻¹⁰ × 10⁵ × cos 60° = -4 × 10⁻⁵ × 0.5 = -2 × 10⁻⁵ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Equipotential Surfaces and Relation Between Field and Potential

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Torque on Current Loop, Magnetic Moment and Galvanometer

A square loop of side \( 0.15 \, \text{m} \) with 50 turns carries \( 1 \, \text{A} \) in a magnetic field of \( 0.6 \,

**Torque on current loop** in magnetic field B is τ = N I A × B, magnitude τ = N I A B sinθ, N turns, I current (A), A area (m²) = l×b for rectangular, θ angle between normal to plane and B. Maximum when plane parallel to B (θ=90°), zero when perpendicular (θ=0°), magnetic moment m = N I A direction along normal via right-hand rule. Torque tau = N I A B sin θ , where A = 0.15 × 0.15 = 0.0225 m² . tau = 50 × 1 × 0.0225 × 0.6 × sin 30° = 0.675 × 0.5 =

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

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Torque on Current Loop, Magnetic Moment and Galvanometer

A rectangular loop of area \( 0.06 \, \text{m}^2 \) with 15 turns carries \( 2.5 \, \text{A} \) in a 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. tau = N I A B sin θ , where θ = 60° to plane means sin 30° with normal. tau = 15 × 2.5 × 0.06 × 0.8 × sin 60° = 1.8 × 0.866 = 1.5588 ≈ 1.56 N m . Using F = q v B sinθ, F =

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

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Solenoid, Toroid and Ampere's Law

What happens to the torque on a rectangular current loop if the magnetic field direction is reversed?

**Solenoid field** inside long solenoid is B = μ₀ n I, n = N/L turns per meter (m⁻¹), uniform and parallel to axis, outside negligible for long solenoid because fields from opposite sides cancel. For n = 1200 m⁻¹, I = 1 A, B = 4π×10⁻⁷×1200 = 1.51×10⁻³ T = 1.51 mT. Torque is given by boldsymboltau = m × B . Reversing the magnetic field B reverses the direction of the torque vector, but its magnitude remains the same. 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 > Solenoid, Toroid and Ampere's Law

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A square loop of side \( 0.12 \, \text{m} \) with 35 turns carries \( 1.8 \, \text{A} \) in a magnetic field of \( 0.5 \

**Biot-Savart law** dB = μ₀/4π·I dl × r̂/r² underlies both straight wire and loop formulas. For square loop side a, area A = a², moment m = N I A, field pattern similar to dipole at large distances, with superposition for N turns. Torque tau = N I A B sin θ , where A = 0.12 × 0.12 = 0.0144 m² . tau = 35 × 1.8 × 0.0144 × 0.5 × sin 30° = 0.9072 × 0.5 = 0.4536 ≈ 0.45 N m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B

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

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A rectangular loop of area \( 0.03 \, \text{m}^2 \) with 20 turns carries \( 4 \, \text{A} \) in a field of \( 0.9 \, \t

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. tau = N I A B sin θ , θ = 90° to plane means sin 0° = 1 with normal. tau = 20 × 4 × 0.03 × 0.9 × 1 = 2.16 N m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π

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

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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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Magnetic Field Due to Current - Straight Wire and Circular Loop

A rectangular loop of area \( 0.05 \, \text{m}^2 \) with 15 turns carries \( 2 \, \text{A} \) in a field of \( 1 \, \tex

**Biot-Savart law** dB = μ₀/4π·I dl × r̂/r² underlies both straight wire and loop formulas. For square loop side a, area A = a², moment m = N I A, field pattern similar to dipole at large distances, with superposition for N turns. tau = N I A B sin θ , θ = 90° to plane means sin 0° = 1 with normal. tau = 15 × 2 × 0.05 × 1 × 1 = 1.5 N m . 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 > Magnetic Field Due to Current - Straight Wire and Circular Loop

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A rectangular loop of area \( 0.08 \, \text{m}^2 \) with 15 turns carries \( 2.5 \, \text{A} \) in a field of \( 0.6 \,

**Biot-Savart law** dB = μ₀/4π·I dl × r̂/r² underlies both straight wire and loop formulas. For square loop side a, area A = a², moment m = N I A, field pattern similar to dipole at large distances, with superposition for N turns. tau = N I A B sin θ , where θ = 30° to plane means sin 60° with normal. tau = 15 × 2.5 × 0.08 × 0.6 × sin 30° = 1.8 × 0.5 = 0.9 N m . 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

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A square loop of side \( 0.2 \, \text{m} \) with 30 turns carries \( 1.5 \, \text{A} \) in a magnetic field of \( 0.4 \,

**Biot-Savart law** dB = μ₀/4π·I dl × r̂/r² underlies both straight wire and loop formulas. For square loop side a, area A = a², moment m = N I A, field pattern similar to dipole at large distances, with superposition for N turns. Torque tau = N I A B sin θ , where A = 0.2 × 0.2 = 0.04 m² . tau = 30 × 1.5 × 0.04 × 0.4 × sin 60° = 1.8 × 0.4 × 0.866 = 0.6235 ≈ 0.62 N m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π

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

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