#force calculation

Force on Current-Carrying Conductor and Between Parallel Wires

Two parallel wires \( 0.08 \, \text{m} \) apart carry \( 7 \, \text{A} \) and \( 4 \, \text{A} \) in opposite directions

**Parallel current interaction** arises because each wire's field B = μ₀ I/(2π d) exerts force F = I l B on other. Force per length f = B I, leading to f = μ₀ I₁ I₂/(2π d). This defines ampere: two wires 1 m apart carrying 1 A exert 2×10⁻⁷ N/m. f = (μ₀ I₁ I₂/2 π d) . f = (4 π × 10⁻⁷ × 7 × 4/2 π × 0.08) = (112 × 10⁻⁷/0.16) = 7 × 10⁻⁶ N/m . 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 > Force on Current-Carrying Conductor and Between Parallel Wires

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Superposition Principle and Equilibrium of Charges

Two point charges \( q_1 = 4 \, \mu\text{C} \) and \( q_2 = -6 \, \mu\text{C} \) are placed 20 cm apart in vacuum. What

**Superposition principle** asserts net Coulomb force on charge equals vector sum of forces from each other charge independently, F_net = Σ F_i, where F_i = k q q_i/r_i² r̂_i. In equilateral triangle or square symmetry, components may cancel at centroid, producing equilibrium. Using Coulomb's law: F = k (|q₁ q₂|/r²) . Given: k = 9 × 10⁹ N·m²/C² , q₁ = 4 × 10⁻⁶ C , q₂ = -6 × 10⁻⁶ C , r = 0.2 m . Magnitude: |q₁ q₂| = (4 × 10⁻⁶) × (6 × 10⁻⁶) = 24 × 10⁻¹² C² . r² = (0.2)² = 0.04 m² . F = 9

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges

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Superposition Principle and Equilibrium of Charges

Two point charges \( 6 \times 10^{-7} \, \text{C} \) and \( 9 \times 10^{-7} \, \text{C} \) are 45 cm apart in air. What

**Independent action of charges** allows total force or field as vector sum. Geometry dictates distances to evaluation point, and resultant follows Σ k q_i/r_i², explaining zero field at symmetric centres for equal charges. Using Coulomb’s law: F = k (|q₁ q₂|/r²) . k = 9 × 10⁹ N·m²/C² , q₁ = 6 × 10⁻⁷ C , q₂ = 9 × 10⁻⁷ C , r = 0.45 m . |q₁ q₂| = 6 × 9 × 10⁻¹⁴ = 54 × 10⁻¹⁴ C² . r² = (0.45)² = 0.2025 m² . F = 9 × 10⁹ × (54 × 10⁻¹⁴/0.2025) = 9 × 10⁹ × 2.667 ×

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges

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Coulomb's Law and Force Between Point Charges

Two point charges \( 7 \times 10^{-7} \, \text{C} \) and \( -2 \times 10^{-7} \, \text{C} \) are 70 cm apart in vacuum.

**Electrostatic force** described by F = (1/4π ε₀)·q₁q₂/r² obeys Newton's third law. Magnitude depends on q₁q₂ and 1/r², enabling quantitative estimation at given separation, with sign indicating attraction or repulsion. Using Coulomb’s law: F = k (|q₁ q₂|/r²) . k = 9 × 10⁹ N·m²/C² , q₁ = 7 × 10⁻⁷ C , q₂ = -2 × 10⁻⁷ C , r = 0.7 m . |q₁ q₂| = 7 × 2 × 10⁻¹⁴ = 14 × 10⁻¹⁴ C² . r² = (0.7)² = 0.49 m² . F = 9 × 10⁹ × (14 × 10⁻¹⁴/0.49) = 9 × 10⁹ × 2.857 × 10⁻¹³ = 0.00257

Ref: NCERT > Physics Book > Electric Charges and Fields > Coulomb's Law and Force Between Point Charges

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Coulomb's Law and Force Between Point Charges

Two point charges \( 4 \times 10^{-7} \, \text{C} \) and \( -8 \times 10^{-7} \, \text{C} \) are 90 cm apart in vacuum.

**Electrostatic force** described by F = (1/4π ε₀)·q₁q₂/r² obeys Newton's third law. Magnitude depends on q₁q₂ and 1/r², enabling quantitative estimation at given separation, with sign indicating attraction or repulsion. Using Coulomb’s law: F = k (|q₁ q₂|/r²) . k = 9 × 10⁹ N·m²/C² , q₁ = 4 × 10⁻⁷ C , q₂ = -8 × 10⁻⁷ C , r = 0.9 m . |q₁ q₂| = 4 × 8 × 10⁻¹⁴ = 32 × 10⁻¹⁴ C² . r² = (0.9)² = 0.81 m² . F = 9 × 10⁹ × (32 × 10⁻¹⁴/0.81) = 9 × 10⁹ × 3.951 × 10⁻¹³ = 0.00356

Ref: NCERT > Physics Book > Electric Charges and Fields > Coulomb's Law and Force Between Point Charges

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PHYSICS

A wire of length 0.6 m carrying 9 A is at 60Â° to a magnetic field of 0.4 T . What is the force on the wire?

Given: A wire of length 0.6 m carrying 9 A is at 60Â° to a magnetic field of 0.4 T . What is the force on the wire? These values define the system as per NCERT data. Formula: F = I l B sin Î¸. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: F = 9 Ã— 0.6 Ã— 0.4 Ã— sin 60Â° = 2.16 Ã— 0.866 = 1.8706 approx 1.87 N . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/sÂ², 10â»âµ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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Surface Tension, Capillarity and Excess Pressure

What is the force on a submarine window (0.06m2) at 400m depth in seawater (ρ\=1.03×103kg/m3), interior at atmospheric p

Gauge pressure: Pg = ρgh = 1.03×103×9.8×400 = 4.0376×106Pa. F = PgA = 4.0376×106×0.06 = 2.42256×105N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.4 × 10⁵ N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Surface Tension, Capillarity and Excess Pressure

A 1500kg car is lifted by a hydraulic system with a small piston of radius 6cm and a large piston of radius 18cm. What f

F1 = A1A2F2, F2 = 1500×9.8 = 14700N. A1 = π(0.06)2, A2 = π(0.18)2, A1A2 = 0.00360.0324 = 19. F1 = 147009 = 1633.33N≈1633N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1633 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.