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#distance calculation

18 public questions tagged with this topic.

A long wire carries \( 12 \, \text{A} \). At what distance is the magnetic field \( 2.4 \times 10^{-6} \, \text{T} \)? (

**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. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 12/2 π × 2.4 × 10⁻⁶) = (48 × 10⁻⁷/4.8 × 10⁻⁶) = 1 m . Using F = q v B sinθ, F = I l B sinθ, B =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

A long wire carries \( 18 \, \text{A} \). At what distance is the magnetic field \( 6 \times 10^{-6} \, \text{T} \)? (\(

**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. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 18/2 π × 6 × 10⁻⁶) = (72 × 10⁻⁷/12 × 10⁻⁶) = 0.6 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 Force on Moving Charge - Lorentz Force and Motion

A long wire carries \( 16 \, \text{A} \). At what distance is the magnetic field \( 8 \times 10^{-6} \, \text{T} \)? (\(

**Lorentz force** on charge q moving with velocity v in magnetic field B is F = q v × B, magnitude F = q v B sinθ, θ angle between v and B (degrees), unit N. Direction perpendicular to both v and B via right-hand rule. When v ⊥ B, motion circular with radius r = m v/(q B), centripetal force provided by magnetic force. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 16/2 π × 8 × 10⁻⁶) = (64 × 10⁻⁷/16 × 10⁻⁶) = 0.4 m . Using F

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

A long wire carries \( 24 \, \text{A} \). At what distance is the magnetic field \( 4.8 \times 10^{-6} \, \text{T} \)? (

**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. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 24/2 π × 4.8 × 10⁻⁶) = (96 × 10⁻⁷/9.6 × 10⁻⁶) = 1 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 Force on Moving Charge - Lorentz Force and Motion

Two particles of masses 4kg and 6kg are at (2,0) and (0,3) respectively. What is the distance of their center of mass fr

X = (4×2)+(6×0)4+6 = 810 = 0.8. Y = (4×0)+(6×3)4+6 = 1810 = 1.8. Distance = (0.8)2+(1.8)2 = 0.64+3.24 = 3.88≈1.97m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.97 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

Two particles of masses 2kg and 8kg are at (1,5) and (4,1) respectively. What is the distance of their center of mass fr

X = (2×1)+(8×4)2+8 = 2+3210 = 3.4. Y = (2×5)+(8×1)2+8 = 10+810 = 1.8. Distance = (3.4)2+(1.8)2 = 11.56+3.24 = 14.8≈3.85m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.85 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

Two particles of masses 6kg and 4kg are at (2,3) and (8,7) respectively. What is the distance of their center of mass fr

X = (6×2)+(4×8)6+4 = 12+3210 = 4.4. Y = (6×3)+(4×7)6+4 = 18+2810 = 4.6. Distance = (4.4)2+(4.6)2 = 19.36+21.16 = 40.52≈6.36m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.36 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

Two particles of masses 7kg and 3kg are at (0,6) and (5,0) respectively. What is the distance of their center of mass fr

X = (7×0)+(3×5)7+3 = 1510 = 1.5. Y = (7×6)+(3×0)7+3 = 4210 = 4.2. Distance = (1.5)2+(4.2)2 = 2.25+17.64 = 19.89≈4.46m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.46 m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A train moving at 90km/h accelerates at 0.8m/s2 for 15s, then decelerates at 2.5m/s2 to rest. What is the total distance

Speed: 90km/h=25m/s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 730 m as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving

A rocket accelerates from rest at 4m/s2 for 6s, then moves at constant speed for 5s. What is the total distance covered?

Phase 1: v=4⋅6=24m/s, x1=12⋅4⋅(6)2=72m. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 192 m as the result, so option D is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving